# NCERT Solutions for Class 12 Maths Chapter 7 Integrals (ex 7.5) Exercise 7.5

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## NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.5) Exercise 7.5

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### Access NCERT Solutions For Class 12 Maths Chapter 7 – Integrals

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### NCERT Solutions For Class 12 Maths Chapter 7 Integrals Exercise 7.5

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Q.1

$\text{Integrate the rational function\hspace{0.17em}}\frac{\mathrm{x}}{\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)}$

Ans

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)}=\frac{\mathrm{A}}{\mathrm{x}+1}+\frac{\mathrm{B}}{\mathrm{x}+2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\mathrm{A}\left(\mathrm{x}+2\right)+\mathrm{B}\left(\mathrm{x}+1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\left(\mathrm{A}+\mathrm{B}\right)\mathrm{x}+2\mathrm{A}+\mathrm{B}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{A}+\mathrm{B}=1\text{and}2\mathrm{A}+\mathrm{B}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{A}=-1\text{and B}=2\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)}=\frac{-1}{\mathrm{x}+1}+\frac{2}{\mathrm{x}+2}\\ \therefore \int \frac{\mathrm{x}}{\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)}\mathrm{dx}=\int \frac{-1}{\mathrm{x}+1}\text{\hspace{0.17em}}\mathrm{dx}+\int \frac{2}{\mathrm{x}+2}\text{\hspace{0.17em}}\mathrm{dx}\\ =\mathrm{log}{\left(\mathrm{x}+1\right)}^{-1}+\mathrm{log}{\left(\mathrm{x}+2\right)}^{2}+\mathrm{C}\\ =\mathrm{log}\frac{{\left(\mathrm{x}+2\right)}^{2}}{\left(\mathrm{x}+1\right)}+\mathrm{C}\end{array}$

Q.2

$\text{Integrate the rational function\hspace{0.17em}}\frac{1}{{\mathrm{x}}^{2}-9}$

Ans

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{{\mathrm{x}}^{2}-9}=\frac{1}{\left(\mathrm{x}+3\right)\left(\mathrm{x}-3\right)}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{A}}{\mathrm{x}+3}+\frac{\mathrm{B}}{\mathrm{x}-3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1=\mathrm{A}\left(\mathrm{x}-3\right)+\mathrm{B}\left(\mathrm{x}+3\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1=\left(\mathrm{A}+\mathrm{B}\right)\mathrm{x}+3\left(-\mathrm{A}+\mathrm{B}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{A}+\mathrm{B}=0\text{and}3\left(-\mathrm{A}+\mathrm{B}\right)=1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{A}=-\frac{1}{6}\text{and B}=\frac{1}{6}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{\left(\mathrm{x}+3\right)\left(\mathrm{x}-3\right)}=\frac{-1}{6\left(\mathrm{x}+3\right)}+\frac{1}{6\left(\mathrm{x}-3\right)}\\ \therefore \int \frac{1}{\left(\mathrm{x}+3\right)\left(\mathrm{x}-3\right)}\mathrm{dx}=\int \frac{-1}{6\left(\mathrm{x}+3\right)}\text{\hspace{0.17em}}\mathrm{dx}+\int \frac{1}{6\left(\mathrm{x}-3\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{1}{6}\mathrm{log}|\mathrm{x}+3|+\frac{1}{6}\mathrm{log}|\mathrm{x}-3|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{6}\mathrm{log}|\frac{\mathrm{x}-3}{\mathrm{x}+3}|+\mathrm{C}\end{array}$

Q.3

$\text{Integrate the rational function\hspace{0.17em}}\frac{3\mathrm{x}-1}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)}$

Ans

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\\ \frac{3\mathrm{x}-1}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)}\\ \text{ }=\frac{\mathrm{A}}{\left(\mathrm{x}-1\right)}+\frac{\mathrm{B}}{\left(\mathrm{x}-2\right)}+\frac{\mathrm{C}}{\left(\mathrm{x}-3\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3\mathrm{x}-1=\mathrm{A}\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)+\mathrm{B}\left(\mathrm{x}-1\right)\left(\mathrm{x}-3\right)+\mathrm{C}\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=1,2\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}3\text{respectively in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{A}=1,\text{B}=-5\text{and C}=\text{4}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{3\mathrm{x}-1}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)}=\frac{1}{\mathrm{x}-1}+\frac{-5}{\mathrm{x}-2}+\frac{4}{\left(\mathrm{x}-3\right)}\\ \therefore \int \frac{3\mathrm{x}-1}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{1}{\mathrm{x}-1}\text{\hspace{0.17em}}\mathrm{dx}-\int \frac{5}{\mathrm{x}-2}\text{\hspace{0.17em}}\mathrm{dx}+\int \frac{4}{\left(\mathrm{x}-3\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}\left|\mathrm{x}-1\right|-5\mathrm{log}\left|\mathrm{x}-2\right|+4\mathrm{log}\left|\mathrm{x}-3\right|+\mathrm{C}\end{array}$

Q.4

$\text{Integrate the rational function\hspace{0.17em}}\frac{\mathrm{x}}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)}$

Ans

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{A}}{\left(\mathrm{x}-1\right)}+\frac{\mathrm{B}}{\left(\mathrm{x}-2\right)}+\frac{\mathrm{C}}{\left(\mathrm{x}-3\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\mathrm{A}\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)+\mathrm{B}\left(\mathrm{x}-1\right)\left(\mathrm{x}-3\right)+\mathrm{C}\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=1,2\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}3\text{respectively in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{A}=\frac{1}{2},\text{B}=-2\text{and C}=\frac{3}{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)}=\frac{1}{2\left(\mathrm{x}-1\right)}+\frac{-2}{\mathrm{x}-2}+\frac{3}{2\left(\mathrm{x}-3\right)}\\ \therefore \int \frac{\mathrm{x}}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{2}\int \frac{1}{\mathrm{x}-1}\text{\hspace{0.17em}}\mathrm{dx}-2\int \frac{1}{\mathrm{x}-2}\text{\hspace{0.17em}}\mathrm{dx}+\frac{3}{2}\int \frac{1}{\left(\mathrm{x}-3\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\mathrm{log}|\mathrm{x}-1|-2\mathrm{log}|\mathrm{x}-2|+\frac{3}{2}\mathrm{log}|\mathrm{x}-3|+\mathrm{C}\end{array}$

Q.5

$\text{Integrate the rational function\hspace{0.17em}}\frac{2\mathrm{x}}{{\mathrm{x}}^{2}+3\mathrm{x}+2}$

Ans

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\frac{2\mathrm{x}}{{\mathrm{x}}^{2}+3\mathrm{x}+2}=\frac{2\mathrm{x}}{\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{A}}{\mathrm{x}+1}+\frac{\mathrm{B}}{\mathrm{x}+2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}=\mathrm{A}\left(\mathrm{x}+2\right)+\mathrm{B}\left(\mathrm{x}+1\right)\text{\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=-1\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}-2\text{respectively in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{A}=-2,\text{B}=4\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{2\mathrm{x}}{\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)}=\frac{-2}{\mathrm{x}+1}+\frac{4}{\mathrm{x}+2}\\ \therefore \int \frac{2\mathrm{x}}{\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{-2}{\mathrm{x}+1}\text{\hspace{0.17em}}\mathrm{dx}+\int \frac{4}{\mathrm{x}+2}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=-2\mathrm{log}|\mathrm{x}+1|+4\mathrm{log}|\mathrm{x}+2|+\mathrm{C}\end{array}$

Q.6

$\text{Integrate the rational function\hspace{0.17em}}\frac{1-{\mathrm{x}}^{2}}{\mathrm{x}\left(1-2\mathrm{x}\right)}$

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em} }\frac{1-{\mathrm{x}}^{2}}{\mathrm{x}\left(1-2\mathrm{x}\right)}=\frac{1}{2}+\frac{1}{2}\left\{\frac{2-\mathrm{x}}{\mathrm{x}\left(1-2\mathrm{x}\right)}\right\}\text{ }...\left(\mathrm{i}\right)\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{2-\mathrm{x}}{\mathrm{x}\left(1-2\mathrm{x}\right)}=\frac{\mathrm{A}}{\mathrm{x}}+\frac{\mathrm{B}}{1-2\mathrm{x}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2-\mathrm{x}=\mathrm{A}\left(1-2\mathrm{x}\right)+\mathrm{Bx}\\ \mathrm{Substituting}\text{x}=0\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}\frac{1}{2}\text{respectively in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{A}=2,\text{B}=3\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{2-\mathrm{x}}{\mathrm{x}\left(1-2\mathrm{x}\right)}=\frac{2}{\mathrm{x}}+\frac{3}{1-2\mathrm{x}}\\ \mathrm{From}\text{​ equation}\left(\mathrm{i}\right),\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{we}\text{​​ have}\\ \int \frac{1-{\mathrm{x}}^{2}}{\mathrm{x}\left(1-2\mathrm{x}\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{1}{2}\text{\hspace{0.17em}}\mathrm{dx}+\frac{1}{2}\int \frac{2}{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}+\frac{1}{2}\int \frac{3}{1-2\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\mathrm{x}+\frac{1}{2}\left(2\mathrm{log}\left|\mathrm{x}\right|-\frac{3}{4}\mathrm{log}\left|1-2\mathrm{x}\right|\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\mathrm{x}+\mathrm{log}\left|\mathrm{x}\right|-\frac{3}{4}\mathrm{log}\left|1-2\mathrm{x}\right|+\mathrm{C}\end{array}$

Q.7

$\text{Integrate the rational function\hspace{0.17em}}\frac{\mathrm{x}}{\left({\mathrm{x}}^{2}+1\right)\left(\mathrm{x}-1\right)}$

Ans

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{\left({\mathrm{x}}^{2}+1\right)\left(\mathrm{x}-1\right)}\text{\hspace{0.17em}}=\frac{\mathrm{Ax}+\mathrm{B}}{{\mathrm{x}}^{2}+1}+\frac{\mathrm{C}}{\mathrm{x}-1}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{x}=\left(\mathrm{Ax}+\mathrm{B}\right)\left(\mathrm{x}-1\right)+\mathrm{C}\left({\mathrm{x}}^{2}+1\right)\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{Ax}}^{2}-\mathrm{Ax}+\mathrm{Bx}-\mathrm{B}+{\mathrm{Cx}}^{2}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{A}+\mathrm{C}\right){\mathrm{x}}^{2}+\left(-\mathrm{A}+\mathrm{B}\right)\mathrm{x}+\left(-\mathrm{B}+\mathrm{C}\right)\\ \mathrm{Equating}{\text{coefficients of x}}^{\text{2}}\text{, x and constant term from both sides,}\\ \text{we get}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{A}+\mathrm{C}=0,\text{}-\text{A}+\text{B}=1\text{and\hspace{0.17em}\hspace{0.17em}}-\mathrm{B}+\mathrm{C}=0\\ \mathrm{On}\text{​ solving these equations, we get}\\ \text{A}=-\frac{1}{2}\text{,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}B}=\frac{1}{2}\text{,\hspace{0.17em}\hspace{0.17em}C}=\frac{1}{2}\\ \mathrm{So},\text{from equation}\left(\mathrm{i}\right)\text{, we get}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{\left({\mathrm{x}}^{2}+1\right)\left(\mathrm{x}-1\right)}\text{\hspace{0.17em}}=\frac{\frac{1}{2}\left(-\mathrm{x}+1\right)}{{\mathrm{x}}^{2}+1}+\frac{\frac{1}{2}}{\mathrm{x}-1}\\ \therefore \int \frac{\mathrm{x}}{\left({\mathrm{x}}^{2}+1\right)\left(\mathrm{x}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{2}\int \frac{\left(-\mathrm{x}\right)}{{\mathrm{x}}^{2}+1}\text{\hspace{0.17em}}\mathrm{dx}+\frac{1}{2}\int \frac{1}{{\mathrm{x}}^{2}+1}\text{\hspace{0.17em}}\mathrm{dx}+\frac{1}{2}\int \frac{1}{\mathrm{x}-1}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=-\frac{1}{4}\mathrm{log}|{\mathrm{x}}^{2}+1|+\frac{1}{2}{\mathrm{tan}}^{-1}\mathrm{x}+\frac{1}{2}\mathrm{log}|\mathrm{x}-1|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\mathrm{log}|\mathrm{x}-1|-\frac{1}{4}\mathrm{log}|{\mathrm{x}}^{2}+1|+\frac{1}{2}{\mathrm{tan}}^{-1}\mathrm{x}+\mathrm{C}\end{array}$

Q.8

$\text{Integrate the rational function\hspace{0.17em}}\frac{\mathrm{x}}{{\left(\mathrm{x}-1\right)}^{2}\left(\mathrm{x}+2\right)}$

Ans

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{{\left(\mathrm{x}-1\right)}^{2}\left(\mathrm{x}+2\right)}=\frac{\mathrm{A}}{\left(\mathrm{x}-1\right)}+\frac{\mathrm{B}}{{\left(\mathrm{x}-1\right)}^{2}}+\frac{\mathrm{C}}{\left(\mathrm{x}+2\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\mathrm{A}\left(\mathrm{x}-1\right)\left(\mathrm{x}+2\right)+\mathrm{B}\left(\mathrm{x}+2\right)+\mathrm{C}{\left(\mathrm{x}-1\right)}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=1\text{in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} B}=\frac{1}{3}\\ \mathrm{Equating}{\text{coefficients of x}}^{\text{2}}\text{and constant term, we get}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}A}+\text{C}=\text{0}\\ -\text{2A}+2\mathrm{B}+\mathrm{C}=0\\ \mathrm{On}\text{solving, we get}\\ \text{A}=\frac{2}{9}\text{and C}=\frac{-2}{9}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{{\left(\mathrm{x}-1\right)}^{2}\left(\mathrm{x}+2\right)}=\frac{2}{9\left(\mathrm{x}-1\right)}+\frac{1}{3{\left(\mathrm{x}-1\right)}^{2}}-\frac{2}{9\left(\mathrm{x}+2\right)}\\ \therefore \int \frac{\mathrm{x}}{{\left(\mathrm{x}-1\right)}^{2}\left(\mathrm{x}+2\right)}\text{\hspace{0.17em}}\mathrm{dx}=\frac{2}{9}\int \frac{1}{\mathrm{x}-1}\text{\hspace{0.17em}}\mathrm{dx}+\frac{1}{3}\int \frac{1}{{\left(\mathrm{x}-1\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}-\frac{2}{9}\int \frac{1}{\left(\mathrm{x}+2\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{2}{9}\mathrm{log}|\mathrm{x}-1|+\frac{1}{3}\frac{-1}{\left(\mathrm{x}-1\right)}-\frac{2}{9}\mathrm{log}|\mathrm{x}+2|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2}{9}\mathrm{log}|\frac{\mathrm{x}-1}{\mathrm{x}+2}|-\frac{1}{3}\frac{1}{\left(\mathrm{x}-1\right)}+\mathrm{C}\end{array}$

Q.9

$\text{Integrate the rational function\hspace{0.17em}}\frac{3\mathrm{x}+5}{{\mathrm{x}}^{3}-{\mathrm{x}}^{2}-\mathrm{x}+1}$

Ans

$\begin{array}{l}\mathrm{Given},\text{\hspace{0.17em}\hspace{0.17em}}\frac{3\mathrm{x}+5}{{\mathrm{x}}^{3}-{\mathrm{x}}^{2}-\mathrm{x}+1}=\frac{3\mathrm{x}+5}{{\mathrm{x}}^{2}\left(\mathrm{x}-1\right)-1\left(\mathrm{x}-1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{3\mathrm{x}+5}{\left(\mathrm{x}-1\right)\left({\mathrm{x}}^{2}-1\right)}\\ \text{\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3\mathrm{x}+5}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-1\right)\left(\mathrm{x}+1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{3\mathrm{x}+5}{{\left(\mathrm{x}-1\right)}^{2}\left(\mathrm{x}+1\right)}\\ \mathrm{Let},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{3\mathrm{x}+5}{{\left(\mathrm{x}-1\right)}^{2}\left(\mathrm{x}+1\right)}=\frac{\mathrm{A}}{\left(\mathrm{x}-1\right)}+\frac{\mathrm{B}}{{\left(\mathrm{x}-1\right)}^{2}}+\frac{\mathrm{C}}{\left(\mathrm{x}+1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3\mathrm{x}+5=\mathrm{A}\left(\mathrm{x}-1\right)\left(\mathrm{x}+1\right)+\mathrm{B}\left(\mathrm{x}+1\right)+\mathrm{C}{\left(\mathrm{x}-1\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\mathrm{A}\left({\mathrm{x}}^{2}-1\right)+\mathrm{B}\left(\mathrm{x}+1\right)+\mathrm{C}\left({\mathrm{x}}^{2}-2\mathrm{x}+1\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=\text{1 in equation}\left(\mathrm{i}\right)\text{, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}B}=\frac{8}{2}=4\\ \mathrm{Equating}{\text{coefficients of x}}^{\text{2}}\text{, x and constant term,we get}\\ \text{A}+\text{C}=\text{0 and B}-2\mathrm{C}=3\\ \mathrm{On}\text{solving these equations, we get}\\ \text{A}=-\frac{1}{2}\text{and C}=\frac{1}{2}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{3\mathrm{x}+5}{{\left(\mathrm{x}-1\right)}^{2}\left(\mathrm{x}+1\right)}=\frac{-1}{2\left(\mathrm{x}-1\right)}+\frac{4}{{\left(\mathrm{x}-1\right)}^{2}}+\frac{1}{2\left(\mathrm{x}+1\right)}\\ ⇒\int \frac{3\mathrm{x}+5}{{\left(\mathrm{x}-1\right)}^{2}\left(\mathrm{x}+1\right)}\text{\hspace{0.17em}}\mathrm{dx}=-\frac{1}{2}\int \frac{1}{\left(\mathrm{x}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}+4\int \frac{1}{{\left(\mathrm{x}-1\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}+\frac{1}{2}\int \frac{1}{\left(\mathrm{x}+1\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{1}{2}\mathrm{log}|\mathrm{x}-1|+4\left(\frac{-1}{\mathrm{x}-1}\right)+\frac{1}{2}\mathrm{log}|\mathrm{x}+1|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{2}\mathrm{log}|\frac{\mathrm{x}+1}{\mathrm{x}-1}|-\frac{4}{\mathrm{x}-1}+\mathrm{C}\end{array}$

Q.10

$\text{Integrate the rational function\hspace{0.17em}}\frac{2\mathrm{x}-3}{\left({\mathrm{x}}^{2}-1\right)\left(2\mathrm{x}+3\right)}$

Ans

$\begin{array}{l}\frac{2\mathrm{x}-3}{\left({\mathrm{x}}^{2}-1\right)\left(2\mathrm{x}+3\right)}=\frac{2\mathrm{x}-3}{\left(\mathrm{x}-1\right)\left(\mathrm{x}+1\right)\left(2\mathrm{x}+3\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{A}}{\left(\mathrm{x}+1\right)}+\frac{\mathrm{B}}{\left(\mathrm{x}-1\right)}+\frac{\mathrm{C}}{\left(2\mathrm{x}+3\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}-3=\mathrm{A}\left(\mathrm{x}-1\right)\left(2\mathrm{x}+3\right)+\mathrm{B}\left(\mathrm{x}-1\right)\left(2\mathrm{x}+3\right)+\mathrm{C}\left(\mathrm{x}-1\right)\left(\mathrm{x}+1\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=-1,1\text{and}-\frac{3}{2}\text{respectively in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{A}=\frac{5}{2}\text{, \hspace{0.17em}B}=-\frac{1}{10}\text{and C}=-\frac{24}{5}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\frac{2\mathrm{x}-3}{\left(\mathrm{x}-1\right)\left(\mathrm{x}+1\right)\left(2\mathrm{x}+3\right)}=\frac{5}{2\left(\mathrm{x}+1\right)}-\frac{1}{10\left(\mathrm{x}-1\right)}-\frac{24}{5\left(2\mathrm{x}+3\right)}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}}\int \frac{2\mathrm{x}-3}{\left({\mathrm{x}}^{2}-1\right)\left(2\mathrm{x}+3\right)}\text{\hspace{0.17em}}\mathrm{dx}=\frac{5}{2}\int \frac{1}{\left(\mathrm{x}+1\right)}\text{\hspace{0.17em}}\mathrm{dx}-\frac{1}{10}\int \frac{1}{\left(\mathrm{x}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\frac{24}{5}\int \frac{1}{\left(2\mathrm{x}+3\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{5}{2}\mathrm{log}|\left(\mathrm{x}+1\right)|-\frac{1}{10}\mathrm{log}|\left(\mathrm{x}-1\right)|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}-\frac{24}{5×2}\mathrm{log}|\left(2\mathrm{x}+3\right)|+\mathrm{D}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{5}{2}\mathrm{log}|\left(\mathrm{x}+1\right)|-\frac{1}{10}\mathrm{log}|\left(\mathrm{x}-1\right)|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\frac{12}{5}\mathrm{log}|\left(2\mathrm{x}+3\right)|+\mathrm{D}\end{array}$

Q.11

$\text{Integrate the rational function\hspace{0.17em}}\frac{5\mathrm{x}}{\left(\mathrm{x}+1\right)\left({\mathrm{x}}^{2}-4\right)}$

Ans

$\begin{array}{l}\frac{5\mathrm{x}}{\left(\mathrm{x}+1\right)\left({\mathrm{x}}^{2}-4\right)}=\frac{5\mathrm{x}}{\left(\mathrm{x}+1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}+2\right)}\\ \text{ \hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{A}}{\left(\mathrm{x}+1\right)}+\frac{\mathrm{B}}{\left(\mathrm{x}+2\right)}+\frac{\mathrm{C}}{\left(\mathrm{x}-2\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}5\mathrm{x}=\mathrm{A}\left(\mathrm{x}-2\right)\left(\mathrm{x}+2\right)+\mathrm{B}\left(\mathrm{x}+1\right)\left(\mathrm{x}-2\right)+\mathrm{C}\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)\text{ }...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=-1,-2\text{and\hspace{0.17em}\hspace{0.17em}}2\text{respectively in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{A}=\frac{5}{3}\text{,\hspace{0.17em}\hspace{0.17em}B}=-\frac{5}{2}\text{and C}=\frac{5}{6}\\ ⇒\text{\hspace{0.17em}}\frac{5\mathrm{x}}{\left(\mathrm{x}+1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}+2\right)}=\frac{5}{3\left(\mathrm{x}+1\right)}-\frac{5}{2\left(\mathrm{x}+2\right)}+\frac{5}{6\left(\mathrm{x}-2\right)}\\ \therefore \int \frac{2\mathrm{x}-3}{\left({\mathrm{x}}^{2}-1\right)\left(2\mathrm{x}+3\right)}\text{\hspace{0.17em}}\mathrm{dx}=\frac{5}{3}\int \frac{1}{\left(\mathrm{x}+1\right)}\text{\hspace{0.17em}}\mathrm{dx}-\frac{5}{2}\int \frac{1}{\left(\mathrm{x}+2\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{ }+\frac{5}{6}\int \frac{1}{\left(\mathrm{x}-2\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{ \hspace{0.17em}\hspace{0.17em}}=\frac{5}{3}\mathrm{log}\left|\left(\mathrm{x}+1\right)\right|-\frac{5}{2}\mathrm{log}\left|\left(\mathrm{x}+2\right)\right|\\ \text{ }+\frac{5}{6}\mathrm{log}\left|\left(\mathrm{x}-2\right)\right|+\mathrm{D}\end{array}$

Q.12

$\text{Integrate the rational function\hspace{0.17em}}\frac{{\mathrm{x}}^{3}+\mathrm{x}+1}{{\mathrm{x}}^{2}-1}$

Ans

$\begin{array}{l}\because \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{3}+\mathrm{x}+1}{{\mathrm{x}}^{2}-1}=\mathrm{x}+\frac{2\mathrm{x}+1}{{\mathrm{x}}^{2}-1}\\ \mathrm{Let},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{2\mathrm{x}+1}{{\mathrm{x}}^{2}-1}=\frac{\mathrm{A}}{\mathrm{x}+1}+\frac{\mathrm{B}}{\mathrm{x}-1}\\ 2\mathrm{x}+1=\mathrm{A}\left(\mathrm{x}-1\right)+\mathrm{B}\left(\mathrm{x}+1\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=-1\text{and 1 respectively in equation\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{i}\right),\text{we get}\\ \text{2}\left(-1\right)+1=\mathrm{A}\left(-1-1\right)+\mathrm{B}\left(-1+1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-1=\mathrm{A}\left(-2\right)⇒\mathrm{A}=\frac{1}{2}\\ \mathrm{and}\text{B}=\frac{3}{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{3}+\mathrm{x}+1}{{\mathrm{x}}^{2}-1}=\mathrm{x}+\frac{1}{2\left(\mathrm{x}+1\right)}+\frac{3}{2\left(\mathrm{x}-1\right)}\\ \int \frac{{\mathrm{x}}^{3}+\mathrm{x}+1}{{\mathrm{x}}^{2}-1}\text{\hspace{0.17em}}\mathrm{dx}=\int \mathrm{x}\text{\hspace{0.17em}}\mathrm{dx}+\frac{1}{2}\int \frac{1}{\left(\mathrm{x}+1\right)}\text{\hspace{0.17em}}\mathrm{dx}+\frac{3}{2}\int \frac{1}{\mathrm{x}-1}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{{\mathrm{x}}^{2}}{2}+\frac{1}{2}\mathrm{log}|\mathrm{x}+1|+\frac{3}{2}\mathrm{log}|\mathrm{x}-1|+\mathrm{C}\end{array}$

Q.13

$\text{Integrate the rational function\hspace{0.17em}}\frac{2}{\left(1-\mathrm{x}\right)\left(1+{\mathrm{x}}^{2}\right)}$

Ans

$\begin{array}{l}\mathrm{We}\text{have,}\frac{2}{\left(1-\mathrm{x}\right)\left(1+{\mathrm{x}}^{2}\right)}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{2}{\left(1-\mathrm{x}\right)\left(1+{\mathrm{x}}^{2}\right)}=\frac{\mathrm{A}}{1-\mathrm{x}}+\frac{\mathrm{Bx}+\mathrm{C}}{1+{\mathrm{x}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2=\mathrm{A}\left(1+{\mathrm{x}}^{2}\right)+\left(\mathrm{Bx}+\mathrm{C}\right)\left(1-\mathrm{x}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2={\mathrm{x}}^{2}\left(\mathrm{A}-\mathrm{B}\right)+\mathrm{x}\left(\mathrm{B}-\mathrm{C}\right)+\left(\mathrm{A}+\mathrm{C}\right)\\ \mathrm{Equating}{\text{the coefficients of x}}^{\text{2}}\text{, x and constant term, we get}\\ \mathrm{A}-\mathrm{B}=0\\ \mathrm{B}-\mathrm{C}=0\\ \mathrm{A}+\mathrm{C}=2\\ \text{On solving these equations, we get}\\ \mathrm{A}=1,\text{}\mathrm{B}=1\text{}\mathrm{and}\text{}\mathrm{C}=1\\ \therefore \frac{2}{\left(1-\mathrm{x}\right)\left(1+{\mathrm{x}}^{2}\right)}=\frac{1}{1-\mathrm{x}}+\frac{\mathrm{x}+1}{1+{\mathrm{x}}^{2}}\\ ⇒\int \frac{2}{\left(1-\mathrm{x}\right)\left(1+{\mathrm{x}}^{2}\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{1}{1-\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}+\int \frac{\mathrm{x}+1}{1+{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \frac{1}{1-\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}+\int \frac{\mathrm{x}}{1+{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}+\int \frac{1}{1+{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\mathrm{log}|1-\mathrm{x}|+\frac{1}{2}\mathrm{log}|1+{\mathrm{x}}^{2}|+{\mathrm{tan}}^{-1}\mathrm{x}+\mathrm{C}\end{array}$

Q.14

$\text{Integrate the rational function}\frac{3\mathrm{x}-1}{{\left(\mathrm{x}+2\right)}^{2}}$

Ans

$\begin{array}{l}\mathrm{We}\text{have,}\frac{3\mathrm{x}-1}{{\left(\mathrm{x}+2\right)}^{2}}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\frac{3\mathrm{x}-1}{{\left(\mathrm{x}+2\right)}^{2}}=\frac{\mathrm{A}}{\left(\mathrm{x}+2\right)}+\frac{\mathrm{B}}{{\left(\mathrm{x}+2\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3\mathrm{x}-1=\mathrm{A}\left(\mathrm{x}+2\right)+\mathrm{B}\\ \mathrm{Equating}{\text{the coefficients of x}}^{\text{2}}\text{, x and constant term, we get}\\ \mathrm{A}=3\\ 2\mathrm{A}+\mathrm{B}=-1\\ \mathrm{Solving},\text{the above equations, we get}\\ \text{A}=3\text{}\mathrm{and}\text{}\mathrm{B}=-7\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{3\mathrm{x}-1}{{\left(\mathrm{x}+2\right)}^{2}}=\frac{3}{\left(\mathrm{x}+2\right)}-\frac{7}{{\left(\mathrm{x}+2\right)}^{2}}\\ ⇒\int \frac{3\mathrm{x}-1}{{\left(\mathrm{x}+2\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{3}{\left(\mathrm{x}+2\right)}\text{\hspace{0.17em}}\mathrm{dx}-\int \frac{7}{{\left(\mathrm{x}+2\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\mathrm{log}|\mathrm{x}+2|-7\left(\frac{-1}{\mathrm{x}+2}\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\mathrm{log}|\mathrm{x}+2|+\frac{7}{\mathrm{x}+2}+\mathrm{C}\end{array}$

Q.15

$\text{Integrate the rational function\hspace{0.17em}}\frac{1}{{\mathrm{x}}^{4}-1}$

Ans

$\begin{array}{l}\mathrm{We}\text{have,}\frac{1}{{\mathrm{x}}^{4}-1}=\frac{1}{\left(\mathrm{x}+1\right)\left(\mathrm{x}-1\right)\left({\mathrm{x}}^{2}+1\right)}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{{\mathrm{x}}^{4}-1}=\frac{\mathrm{A}}{\left(\mathrm{x}+1\right)}+\frac{\mathrm{B}}{\left(\mathrm{x}-1\right)}+\frac{\mathrm{Cx}+\mathrm{D}}{\left({\mathrm{x}}^{2}+1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1=\mathrm{A}\left(\mathrm{x}-1\right)\left({\mathrm{x}}^{2}+1\right)+\mathrm{B}\left(\mathrm{x}+1\right)\left({\mathrm{x}}^{2}+1\right)+\left({\mathrm{x}}^{2}-1\right)\left(\mathrm{Cx}+\mathrm{D}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1=\mathrm{A}\left({\mathrm{x}}^{3}+\mathrm{x}-{\mathrm{x}}^{2}-1\right)+\mathrm{B}\left({\mathrm{x}}^{3}+\mathrm{x}+{\mathrm{x}}^{2}+1\right)+\mathrm{C}\left({\mathrm{x}}^{3}-\mathrm{x}\right)+\mathrm{D}\left({\mathrm{x}}^{2}-1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}1={\mathrm{x}}^{3}\left(\mathrm{A}+\mathrm{B}+\mathrm{C}\right)+{\mathrm{x}}^{2}\left(-\mathrm{A}+\mathrm{B}+\mathrm{D}\right)+\mathrm{x}\left(\mathrm{A}+\mathrm{B}-\mathrm{C}\right)+\left(-\mathrm{A}+\mathrm{B}-\mathrm{D}\right)\\ \mathrm{Equating}{\text{the coefficients of x}}^{\text{2}}\text{, x and constant term, we get}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{A}+\mathrm{B}+\mathrm{C}=0\\ \text{\hspace{0.17em}}-\mathrm{A}+\mathrm{B}+\mathrm{D}=0\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{A}+\mathrm{B}-\mathrm{C}=0\\ \text{\hspace{0.17em}}-\mathrm{A}+\mathrm{B}-\mathrm{D}=1\\ \mathrm{Solving},\text{the above equations, we get}\\ \text{A}=-\frac{1}{4},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{B}=\frac{1}{4},\text{\hspace{0.17em}}\mathrm{C}=0\text{and D}=-\frac{1}{2}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{{\mathrm{x}}^{4}-1}=\frac{-1}{4\left(\mathrm{x}+1\right)}+\frac{1}{4\left(\mathrm{x}-1\right)}+\frac{-1}{2\left({\mathrm{x}}^{2}+1\right)}\\ ⇒\int \frac{1}{{\mathrm{x}}^{4}-1}\text{\hspace{0.17em}}\mathrm{dx}=-\frac{1}{4}\int \frac{1}{\mathrm{x}+1}\text{\hspace{0.17em}}\mathrm{dx}+\frac{1}{4}\int \frac{1}{\mathrm{x}-1}\text{\hspace{0.17em}}\mathrm{dx}-\frac{1}{2}\int \frac{1}{{\mathrm{x}}^{2}+1}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=-\frac{1}{4}\mathrm{log}|\left(\mathrm{x}+1\right)|+\frac{1}{4}\mathrm{log}|\left(\mathrm{x}-1\right)|-\frac{1}{2}{\mathrm{tan}}^{-1}\mathrm{x}+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{1}{4}\mathrm{log}|\frac{\mathrm{x}-1}{\mathrm{x}+1}|-\frac{1}{2}{\mathrm{tan}}^{-1}\mathrm{x}+\mathrm{C}\end{array}$

Q.16

$\text{Integrate the rational function\hspace{0.17em}}\frac{1}{\mathrm{x}\left({\mathrm{x}}^{\mathrm{n}}-1\right)}$

Ans

$\begin{array}{l}\mathrm{We}\text{have,}\frac{1}{\mathrm{x}\left({\mathrm{x}}^{\mathrm{n}}-1\right)}=\frac{1}{\mathrm{x}\left({\mathrm{x}}^{\mathrm{n}}-1\right)}×\frac{{\mathrm{x}}^{\mathrm{n}-1}}{{\mathrm{x}}^{\mathrm{n}-1}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{x}}^{\mathrm{n}-1}}{{\mathrm{x}}^{\mathrm{n}}\left({\mathrm{x}}^{\mathrm{n}}-1\right)}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{t}={\mathrm{x}}^{\mathrm{n}}⇒\frac{\mathrm{dt}}{\mathrm{dx}}={\mathrm{nx}}^{\mathrm{n}-1}⇒\frac{\mathrm{dt}}{{\mathrm{nx}}^{\mathrm{n}-1}}=\mathrm{dx}\\ \therefore \int \frac{1}{\mathrm{x}\left({\mathrm{x}}^{\mathrm{n}}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{{\mathrm{x}}^{\mathrm{n}-1}}{{\mathrm{x}}^{\mathrm{n}}\left({\mathrm{x}}^{\mathrm{n}}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{1}{\mathrm{n}}\int \frac{1}{\mathrm{t}\left(\mathrm{t}-1\right)}\text{\hspace{0.17em}}\mathrm{dt}\\ \mathrm{Let}\text{}\frac{1}{\mathrm{t}\left(\mathrm{t}-1\right)}=\frac{\mathrm{A}}{\mathrm{t}}+\frac{\mathrm{B}}{\mathrm{t}-1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}1=\mathrm{A}\left(\mathrm{t}-1\right)+\mathrm{Bt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{t}\left(\mathrm{A}+\mathrm{B}\right)-\mathrm{A}\\ \mathrm{Substituting}\text{​ t}=\text{0 and 1 respectively, we get}\\ \mathrm{A}=1\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{B}=-1\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{\mathrm{t}\left(\mathrm{t}-1\right)}=\frac{1}{\mathrm{t}}+\frac{-1}{\mathrm{t}-1}\\ ⇒\int \frac{1}{\mathrm{x}\left({\mathrm{x}}^{\mathrm{n}}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{\mathrm{n}}\int \left(\frac{1}{\mathrm{t}}-\frac{1}{\mathrm{t}-1}\right)\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\mathrm{n}}\int \frac{1}{\mathrm{t}}\text{\hspace{0.17em}}\mathrm{dt}-\frac{1}{\mathrm{n}}\int \frac{1}{\mathrm{t}-1}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\mathrm{n}}\mathrm{log}|\mathrm{t}|-\frac{1}{\mathrm{n}}\mathrm{log}|\mathrm{t}-1|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{\mathrm{n}}\mathrm{log}|{\mathrm{x}}^{\mathrm{n}}|-\frac{1}{\mathrm{n}}\mathrm{log}|{\mathrm{x}}^{\mathrm{n}}-1|+\mathrm{C}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\mathrm{n}}\mathrm{log}|\frac{{\mathrm{x}}^{\mathrm{n}}}{{\mathrm{x}}^{\mathrm{n}}-1}|+\mathrm{C}\end{array}$

Q.17

$\text{Integrate the rational function\hspace{0.17em}}\frac{\mathrm{cosx}}{\left(1-\mathrm{sinx}\right)\left(2-\mathrm{sinx}\right)}$

Ans

$\begin{array}{l}\mathrm{We}\text{have,}\frac{\mathrm{cosx}}{\left(1-\mathrm{sinx}\right)\left(2-\mathrm{sinx}\right)}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{t}=\mathrm{sinx}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=\mathrm{cosx}⇒\frac{\mathrm{dt}}{\mathrm{cosx}}=\mathrm{dx}\\ \therefore \int \frac{\mathrm{cosx}}{\left(1-\mathrm{sinx}\right)\left(2-\mathrm{sinx}\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{\mathrm{cosx}}{\left(1-\mathrm{t}\right)\left(2-\mathrm{t}\right)}\text{\hspace{0.17em}}\frac{\mathrm{dt}}{\mathrm{cosx}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\int \frac{1}{\left(1-\mathrm{t}\right)\left(2-\mathrm{t}\right)}\text{\hspace{0.17em}}\mathrm{dt}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{\left(1-\mathrm{t}\right)\left(2-\mathrm{t}\right)}=\frac{\mathrm{A}}{1-\mathrm{t}}+\frac{\mathrm{B}}{2-\mathrm{t}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}1=\mathrm{A}\left(2-\mathrm{t}\right)+\mathrm{B}\left(1-\mathrm{t}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(1\right)\\ \mathrm{Substituting}\text{t}=1\text{and 2 respectively in eqution}\left(1\right),\text{\hspace{0.17em}}\mathrm{we}\text{get}\\ \text{A}=\text{1 and B}=-1\\ \therefore \frac{1}{\left(1-\mathrm{t}\right)\left(2-\mathrm{t}\right)}=\frac{1}{1-\mathrm{t}}-\frac{1}{2-\mathrm{t}}\\ \mathrm{So},\\ \int \frac{\mathrm{cosx}}{\left(1-\mathrm{sinx}\right)\left(2-\mathrm{sinx}\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{1}{\left(1-\mathrm{t}\right)\left(2-\mathrm{t}\right)}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\int \frac{1}{1-\mathrm{t}}\text{\hspace{0.17em}}\mathrm{dt}-\int \frac{1}{2-\mathrm{t}}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\mathrm{log}|1-\mathrm{t}|+\mathrm{log}|2-\mathrm{t}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\mathrm{log}|\frac{2-\mathrm{t}}{1-\mathrm{t}}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}|\frac{2-\mathrm{sinx}}{1-\mathrm{sinx}}|+\mathrm{C}\end{array}$

Q.18

$\text{Integrate the rational function\hspace{0.17em}}\frac{\left({\mathrm{x}}^{2}+1\right)\left({\mathrm{x}}^{2}+2\right)}{\left({\mathrm{x}}^{2}+3\right)\left({\mathrm{x}}^{2}+4\right)}$

Ans

$\begin{array}{l}\mathrm{We}\text{have,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}}\frac{\left({\mathrm{x}}^{2}+1\right)\left({\mathrm{x}}^{2}+2\right)}{\left({\mathrm{x}}^{2}+3\right)\left({\mathrm{x}}^{2}+4\right)}=1+\frac{\left(4{\mathrm{x}}^{2}+10\right)}{\left({\mathrm{x}}^{2}+3\right)\left({\mathrm{x}}^{2}+4\right)}\\ \mathrm{Let}\text{\hspace{0.17em}},\text{\hspace{0.17em}}\\ \frac{\left(4{\mathrm{x}}^{2}+10\right)}{\left({\mathrm{x}}^{2}+3\right)\left({\mathrm{x}}^{2}+4\right)}=\frac{\mathrm{Ax}+\mathrm{B}}{\left({\mathrm{x}}^{2}+3\right)}+\frac{\mathrm{Cx}+\mathrm{D}}{\left({\mathrm{x}}^{2}+4\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4{\mathrm{x}}^{2}+10\text{\hspace{0.17em}}=\left(\mathrm{Ax}+\mathrm{B}\right)\left({\mathrm{x}}^{2}+4\right)+\left(\mathrm{Cx}+\mathrm{D}\right)\left({\mathrm{x}}^{2}+3\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{3}\left(\mathrm{A}+\mathrm{C}\right)+{\mathrm{x}}^{2}\left(\mathrm{B}+\mathrm{D}\right)+\mathrm{x}\left(4\mathrm{A}+3\mathrm{C}\right)+\left(4\mathrm{B}+3\mathrm{D}\right)\\ \mathrm{Comparing}{\text{the coefficients of x}}^{\text{3}}{\text{, x}}^{\text{2}}\text{, x and constant terms,}\\ \text{we get\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \mathrm{A}+\mathrm{C}=0\\ \mathrm{B}+\mathrm{D}=4\\ 4\mathrm{A}+3\mathrm{C}=0\\ 4\mathrm{B}+3\mathrm{D}=10\\ \mathrm{On}\text{​ solving these equations, we get}\\ \text{A}=0,\mathrm{B}=-2,\mathrm{C}=0\text{and}\mathrm{D}=6\\ \therefore \frac{\left(4{\mathrm{x}}^{2}+10\right)}{\left({\mathrm{x}}^{2}+3\right)\left({\mathrm{x}}^{2}+4\right)}=\frac{-2}{\left({\mathrm{x}}^{2}+3\right)}+\frac{6}{\left({\mathrm{x}}^{2}+4\right)}\\ \mathrm{Then},\\ \int \frac{\left({\mathrm{x}}^{2}+1\right)\left({\mathrm{x}}^{2}+2\right)}{\left({\mathrm{x}}^{2}+3\right)\left({\mathrm{x}}^{2}+4\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int 1\text{\hspace{0.17em}}\mathrm{dx}-\left(\int \frac{-2}{\left({\mathrm{x}}^{2}+3\right)}\text{\hspace{0.17em}}\mathrm{dx}+6\int \frac{1}{\left({\mathrm{x}}^{2}+4\right)}\text{\hspace{0.17em}}\mathrm{dx}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\mathrm{x}+\frac{2}{\sqrt{3}}{\mathrm{tan}}^{-1}\left(\frac{\mathrm{x}}{\sqrt{3}}\right)-\frac{6}{2}{\mathrm{tan}}^{-1}\left(\frac{\mathrm{x}}{2}\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\mathrm{x}+\frac{2}{\sqrt{3}}{\mathrm{tan}}^{-1}\left(\frac{\mathrm{x}}{\sqrt{3}}\right)-3{\mathrm{tan}}^{-1}\left(\frac{\mathrm{x}}{2}\right)+\mathrm{C}\end{array}$

Q.19

$\text{Integrate the rational function\hspace{0.17em}}\frac{2\mathrm{x}}{\left({\mathrm{x}}^{2}+1\right)\left({\mathrm{x}}^{2}+3\right)}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have}, \mathrm{ }\\ \mathrm{ }\frac{2\mathrm{x}}{\left({\mathrm{x}}^{2}+1\right)\left({\mathrm{x}}^{2}+3\right)}\\ \mathrm{Let} \mathrm{t}={\mathrm{x}}^{2}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=2\mathrm{x}⇒\mathrm{dx}=\frac{\mathrm{dt}}{2\mathrm{x}}\\ \int \frac{2\mathrm{x}}{\left({\mathrm{x}}^{2}+1\right)\left({\mathrm{x}}^{2}+3\right)}\mathrm{ }\mathrm{dx}=\int \frac{2\mathrm{x}}{\left(\mathrm{t}+1\right)\left(\mathrm{t}+3\right)}\mathrm{ }\frac{\mathrm{dt}}{2\mathrm{x}}\\ \mathrm{ }=\int \frac{1}{\left(\mathrm{t}+1\right)\left(\mathrm{t}+3\right)}\mathrm{ }\mathrm{dt}\\ \mathrm{Let} \frac{1}{\left(\mathrm{t}+1\right)\left(\mathrm{t}+3\right)}=\frac{\mathrm{A}}{\mathrm{t}+1}+\frac{\mathrm{B}}{\mathrm{t}+3}\\ 1=\mathrm{A}\left(\mathrm{t}+3\right)+\mathrm{B}\left(\mathrm{t}+1\right)\\ \mathrm{Substituting}\mathrm{t}=-1\mathrm{and}-3\mathrm{respectively},\mathrm{we}\mathrm{get}\\ \mathrm{A}=\frac{1}{2} \mathrm{and} \mathrm{B}=-\frac{1}{2}\\ \therefore \frac{1}{\left(\mathrm{t}+1\right)\left(\mathrm{t}+3\right)}=\frac{1}{2\left(\mathrm{t}+1\right)}-\frac{1}{2\left(\mathrm{t}+3\right)}\\ ⇒\int \frac{1}{\left(\mathrm{t}+1\right)\left(\mathrm{t}+3\right)}\mathrm{ }\mathrm{dx}=\int \frac{1}{2\left(\mathrm{t}+1\right)}\mathrm{ }\mathrm{dt}-\int \frac{1}{2\left(\mathrm{t}+3\right)}\mathrm{ }\mathrm{dx}\\ \mathrm{ }=\frac{1}{2}\mathrm{log}|\mathrm{t}+1|-\frac{1}{2}\mathrm{log}|\mathrm{t}+3|+\mathrm{C}\\ \mathrm{ }=\frac{1}{2}\mathrm{log}|\frac{\mathrm{t}+1}{\mathrm{t}+3}|+\mathrm{C}\\ =\frac{1}{2}\mathrm{log}|\frac{{\mathrm{x}}^{2}+1}{{\mathrm{x}}^{2}+3}|+\mathrm{C}\end{array}$

Q.20

$\text{Integrate the rational function\hspace{0.17em}}\frac{1}{\mathrm{x}\left({\mathrm{x}}^{4}-1\right)}$

Ans

$\begin{array}{l}\mathrm{We}\text{have,}\frac{1}{\mathrm{x}\left({\mathrm{x}}^{4}-1\right)}=\frac{1}{\mathrm{x}\left({\mathrm{x}}^{4}-1\right)}×\frac{{\mathrm{x}}^{3}}{{\mathrm{x}}^{3}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{x}}^{3}}{{\mathrm{x}}^{4}\left({\mathrm{x}}^{4}-1\right)}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{t}={\mathrm{x}}^{4}⇒\frac{\mathrm{dt}}{\mathrm{dx}}=4{\mathrm{x}}^{4-1}⇒\frac{\mathrm{dt}}{4{\mathrm{x}}^{3}}=\mathrm{dx}\\ \therefore \int \frac{1}{\mathrm{x}\left({\mathrm{x}}^{4}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{{\mathrm{x}}^{3}}{{\mathrm{x}}^{4}\left({\mathrm{x}}^{4}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\int \frac{{\mathrm{x}}^{3}}{\mathrm{t}\left(\mathrm{t}-1\right)}\text{\hspace{0.17em}}\frac{\mathrm{dt}}{4{\mathrm{x}}^{3}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{4}\int \frac{1}{\mathrm{t}\left(\mathrm{t}-1\right)}\text{\hspace{0.17em}}\mathrm{dt}\\ \mathrm{Let}\text{}\frac{1}{\mathrm{t}\left(\mathrm{t}-1\right)}=\frac{\mathrm{A}}{\mathrm{t}}+\frac{\mathrm{B}}{\mathrm{t}-1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}1=\mathrm{A}\left(\mathrm{t}-1\right)+\mathrm{Bt}\\ \mathrm{Substituting}\text{​ t}=\text{0 and 1 respectively, we get}\\ \mathrm{A}=-1\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{B}=1\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{\mathrm{t}\left(\mathrm{t}-1\right)}=\frac{-1}{\mathrm{t}}+\frac{1}{\mathrm{t}-1}\\ ⇒\int \frac{1}{\mathrm{x}\left({\mathrm{x}}^{4}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}=\frac{1}{4}\int \left(-\frac{1}{\mathrm{t}}+\frac{1}{\mathrm{t}-1}\right)\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=-\frac{1}{4}\int \frac{1}{\mathrm{t}}\text{\hspace{0.17em}}\mathrm{dt}+\frac{1}{4}\int \frac{1}{\mathrm{t}-1}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{1}{4}\mathrm{log}|\mathrm{t}|+\frac{1}{4}\mathrm{log}|\mathrm{t}-1|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=-\frac{1}{4}\mathrm{log}|{\mathrm{x}}^{4}|+\frac{1}{4}\mathrm{log}|{\mathrm{x}}^{4}-1|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{4}\mathrm{log}|\frac{{\mathrm{x}}^{4}-1}{{\mathrm{x}}^{4}}|+\mathrm{C}\end{array}$

Q.21

$\text{Integrate the rational function\hspace{0.17em}}\frac{1}{\left({\mathrm{e}}^{\mathrm{x}}-1\right)}$

Ans

$\begin{array}{l}\int \frac{1}{\left({\mathrm{e}}^{\mathrm{x}}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{t}={\mathrm{e}}^{\mathrm{x}}⇒\frac{\mathrm{dt}}{\mathrm{dx}}={\mathrm{e}}^{\mathrm{x}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{dx}=\frac{\mathrm{dt}}{{\mathrm{e}}^{\mathrm{x}}}=\frac{\mathrm{dt}}{\mathrm{t}}\\ \therefore \int \frac{1}{\left({\mathrm{e}}^{\mathrm{x}}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{1}{\left(\mathrm{t}-1\right)}\text{\hspace{0.17em}}\frac{\mathrm{dt}}{\mathrm{t}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\int \frac{1}{\mathrm{t}\left(\mathrm{t}-1\right)}\mathrm{dt}\\ \mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{\mathrm{t}\left(\mathrm{t}-1\right)}=\frac{\mathrm{A}}{\mathrm{t}}+\frac{\mathrm{B}}{\mathrm{t}-1}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{A}\left(\mathrm{t}-1\right)+\mathrm{Bt}}{\mathrm{t}\left(\mathrm{t}-1\right)}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}1=\mathrm{A}\left(\mathrm{t}-1\right)+\mathrm{Bt}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Putting}\text{t}=0\text{​ and 1 respectively in equation}\left(\mathrm{i}\right),\text{\hspace{0.17em}}\mathrm{we}\text{get}\\ \text{A}=-1\text{and B}=1\\ \therefore \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\frac{1}{\mathrm{t}\left(\mathrm{t}-1\right)}=\frac{-1}{\mathrm{t}}+\frac{1}{\mathrm{t}-1}\\ ⇒\int \frac{1}{\mathrm{t}\left(\mathrm{t}-1\right)}\text{\hspace{0.17em}}\mathrm{dt}=\int \frac{-1}{\mathrm{t}}\text{\hspace{0.17em}}\mathrm{dt}+\int \frac{1}{\mathrm{t}-1}\text{\hspace{0.17em}}\mathrm{dt}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\mathrm{log}|\mathrm{t}|+\mathrm{log}|\mathrm{t}-1|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}|\frac{\mathrm{t}-1}{\mathrm{t}}|+\mathrm{C}\\ \therefore \int \frac{1}{\left({\mathrm{e}}^{\mathrm{x}}-1\right)}\text{\hspace{0.17em}}\mathrm{dx}=\mathrm{log}|\frac{{\mathrm{e}}^{\mathrm{x}}-1}{{\mathrm{e}}^{\mathrm{x}}}|+\mathrm{C}\end{array}$

Q.22

$\begin{array}{l}\mathrm{Choose}\text{ }\mathrm{the}\text{ }\mathrm{correct}\text{ }\mathrm{answer}\\ \int \frac{\mathrm{xdx}}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)}\text{ }\mathrm{dx}\text{ }\mathrm{equals}\\ \left(\mathrm{A}\right)\text{​ }\mathrm{log}\left|\frac{{\left(\mathrm{x}-1\right)}^{2}}{\mathrm{x}-2}\right|+\mathrm{C}\\ \left(\mathrm{B}\right)\text{ }\mathrm{log}\left|\frac{{\left(\mathrm{x}-2\right)}^{2}}{\mathrm{x}-1}\right|+\mathrm{C}\\ \left(\mathrm{C}\right)\text{ }\mathrm{log}\left|{\left(\frac{\mathrm{x}-1}{\mathrm{x}-2}\right)}^{2}\right|+\mathrm{C}\\ \left(\mathrm{D}\right)\text{ }\mathrm{log}\left|\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\right|+\mathrm{C}\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\text{have,}\int \frac{\mathrm{xdx}}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Let}\text{}\frac{\mathrm{x}}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)}=\frac{\mathrm{A}}{\mathrm{x}-1}+\frac{\mathrm{B}}{\mathrm{x}-2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{A}\left(\mathrm{x}-2\right)+\mathrm{B}\left(\mathrm{x}-1\right)}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\mathrm{A}\left(\mathrm{x}-2\right)+\mathrm{B}\left(\mathrm{x}-1\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\text{x}=1\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}2,\text{in equation}\left(\mathrm{i}\right),\text{​\hspace{0.17em}we get}\\ \text{A}=-\text{1 and B}=2\\ ⇒\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)}=\frac{-1}{\mathrm{x}-1}+\frac{2}{\mathrm{x}-2}\\ \therefore \int \frac{\mathrm{x}}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{-1}{\mathrm{x}-1}\text{\hspace{0.17em}}\mathrm{dx}+\int \frac{2}{\mathrm{x}-2}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=-\text{log}|\mathrm{x}-1|+2\mathrm{log}|\mathrm{x}-2|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{log}|\frac{{\left(\mathrm{x}-2\right)}^{2}}{\left(\mathrm{x}-1\right)}|+\mathrm{C}\\ \mathrm{Hence},\text{the option}\left(\text{B}\right)\text{is correct.}\end{array}$

Q.23

$\begin{array}{l}\mathrm{Choose}\text{ }\mathrm{the}\text{ }\mathrm{correct}\text{ }\mathrm{answer}\\ \int \frac{\mathrm{dx}}{\mathrm{x}\left({\mathrm{x}}^{2}+1\right)}\mathrm{equals}\\ \left(\mathrm{A}\right)\text{​ }\mathrm{log}\left|\mathrm{x}\right|-\frac{1}{2}\mathrm{log}\left({\mathrm{x}}^{2}+1\right)+\mathrm{C}\\ \left(\mathrm{B}\right)\text{ }\mathrm{log}\left|\mathrm{x}\right|+\frac{1}{2}\mathrm{log}\left({\mathrm{x}}^{2}+1\right)+\mathrm{C}\\ \left(\mathrm{C}\right)\text{ }-\mathrm{log}\left|\mathrm{x}\right|+\frac{1}{2}\mathrm{log}\left({\mathrm{x}}^{2}+1\right)+\mathrm{C}\\ \left(\mathrm{D}\right)\text{ }\frac{1}{2}\mathrm{log}\left|\mathrm{x}\right|+\mathrm{log}\left({\mathrm{x}}^{2}+1\right)+\mathrm{C}\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\text{have,}\int \frac{\mathrm{dx}}{\mathrm{x}\left({\mathrm{x}}^{2}+1\right)}\\ \mathrm{Let}\text{}\frac{1}{\mathrm{x}\left({\mathrm{x}}^{2}+1\right)}=\frac{\mathrm{A}}{\mathrm{x}}+\frac{\mathrm{Bx}+\mathrm{C}}{\left({\mathrm{x}}^{2}+1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{A}\left({\mathrm{x}}^{2}+1\right)+\left(\mathrm{Bx}+\mathrm{C}\right)\mathrm{x}}{\mathrm{x}\left({\mathrm{x}}^{2}+1\right)}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1=\mathrm{A}\left({\mathrm{x}}^{2}+1\right)+\left({\mathrm{Bx}}^{2}+\mathrm{Cx}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}1={\mathrm{x}}^{2}\left(\mathrm{A}+\mathrm{B}\right)+\mathrm{Cx}+\mathrm{A}\\ \mathrm{Equating}{\text{the coefficients of x}}^{\text{2}}\text{, x and constant term, we get}\\ \mathrm{A}+\mathrm{B}=0,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{C}=0\text{\hspace{0.17em}and A}=1\\ \mathrm{On}\text{solving these equations, we get}\\ \text{A}=\text{1,\hspace{0.17em} B}=-\text{1 and C}=0\\ ⇒\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{\mathrm{x}\left({\mathrm{x}}^{2}+1\right)}=\frac{1}{\mathrm{x}}+\frac{-\mathrm{x}}{\left({\mathrm{x}}^{2}+1\right)}\\ \therefore \int \frac{1}{\mathrm{x}\left({\mathrm{x}}^{2}+1\right)}\text{\hspace{0.17em}}\mathrm{dx}=\int \frac{1}{\mathrm{x}}\text{\hspace{0.17em}}\mathrm{dx}-\int \frac{\mathrm{x}}{\left({\mathrm{x}}^{2}+1\right)}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\text{log}|\mathrm{x}|-2\mathrm{log}|{\mathrm{x}}^{2}+1|+\mathrm{C}\\ \mathrm{Thus},\text{the correct option is}\left(\mathrm{A}\right).\end{array}$