# NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.7) Exercise 7.7

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## NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.7) Exercise 7.7

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### Access NCERT Solutions for Class 12 Chapter 7 – Integrals

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### NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.7

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Q.1

$\begin{array}{l}\begin{array}{l}\mathrm{Choose}\text{}\mathrm{the}\text{}\mathrm{correct}\text{}\mathrm{answer}\\ \int \sqrt{1+{\text{x}}^{2}}\text{}\mathrm{dx}\mathrm{is}\mathrm{equal}\mathrm{to}\end{array}\\ \left(\text{A}\right)\text{}\frac{\text{x}}{2}\sqrt{1+{\text{x}}^{2}}+\frac{1}{2}\mathrm{log}\left|\left(\text{x}+\sqrt{1+{\text{x}}^{2}}\right)\right|+\text{C}\\ \left(\text{B}\right)\text{}\frac{2}{3}{\left(1+{\text{x}}^{2}\right)}^{\frac{3}{2}}+\text{C}\\ \left(\text{C}\right)\text{}\frac{2}{3}\text{x}{\left(1+{\text{x}}^{2}\right)}^{\frac{3}{2}}+\text{C}\\ \left(\text{D}\right)\text{}\frac{{\text{x}}^{2}}{2}\sqrt{1+{\text{x}}^{2}}\text{}+\frac{1}{2}{\text{x}}^{2}\text{}\mathrm{log}\left|\left(\text{x}+\sqrt{1+{\text{x}}^{2}}\right)\right|+\text{C}\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\int \sqrt{1+{\mathrm{x}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\int \sqrt{{\mathrm{x}}^{2}+{1}^{2}}\text{\hspace{0.17em}}\mathrm{dx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[µ\int \sqrt{{\mathrm{x}}^{2}+{\mathrm{a}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^{2}+{\mathrm{a}}^{2}}+\frac{{\mathrm{a}}^{2}}{2}\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}+{\mathrm{a}}^{2}}|\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^{2}+{1}^{2}}+\frac{{1}^{2}}{2}\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}+{1}^{2}}|\right)+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^{2}+1}+\frac{1}{2}\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}+1}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{x}}{2}\sqrt{1+{\mathrm{x}}^{2}}+\frac{1}{2}\mathrm{log}|\mathrm{x}+\sqrt{1+{\mathrm{x}}^{2}}|+\mathrm{C}\\ \mathrm{Hence},\text{â€‹\hspace{0.17em}\hspace{0.17em}}\mathrm{the}\text{correct option is A.}\end{array}$

Q.2

$\begin{array}{l}\begin{array}{l}\mathrm{Choose}\text{}\mathrm{the}\text{}\mathrm{correct}\text{}\mathrm{answer}\\ \int \sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}\text{}\mathrm{dx}\mathrm{is}\mathrm{equal}\mathrm{to}\end{array}\\ \left(\mathrm{A}\right)\text{}\frac{1}{2}\left(\mathrm{x}-4\right)\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}+9\text{}\mathrm{log}\left|\mathrm{x}-4+\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}\right|+\mathrm{C}\\ \left(\mathrm{B}\right)\text{}\frac{1}{4}\left(\mathrm{x}+4\right)\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}+9\mathrm{log}\left|\mathrm{x}+4+\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}\right|+\mathrm{C}\\ \left(\mathrm{C}\right)\text{}\frac{1}{2}\left(\mathrm{x}-4\right)\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}-3\sqrt{2}\text{}\mathrm{log}\left|\mathrm{x}-4+\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}\right|+\mathrm{C}\\ \left(\mathrm{D}\right)\text{}\frac{1}{2}\left(\mathrm{x}-4\right)\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}-\frac{9}{2}\mathrm{log}\left|\mathrm{x}-4+\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}\right|+\mathrm{C}\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\text{have,}\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}=\sqrt{{\left(\mathrm{x}-4\right)}^{2}-9}\\ \mathrm{Let}\text{I}=\int \sqrt{{\left(\mathrm{x}-4\right)}^{2}-{\left(3\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}}=\int \sqrt{{\mathrm{t}}^{2}-{\left(3\right)}^{2}}\text{\hspace{0.17em}}\mathrm{dt}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Let}\text{\hspace{0.17em}t}=\mathrm{x}-4⇒\frac{\mathrm{dt}}{\mathrm{dx}}=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}}\left[µ\int \sqrt{{\mathrm{x}}^{2}-{\mathrm{a}}^{2}}\text{\hspace{0.17em}}\mathrm{dx}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^{2}-{\mathrm{a}}^{2}}-\frac{{\mathrm{a}}^{2}}{2}\mathrm{log}|\mathrm{x}+\sqrt{{\mathrm{x}}^{2}-{\mathrm{a}}^{2}}|\right]\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{t}}{2}\sqrt{{\mathrm{t}}^{2}-{\left(3\right)}^{2}}-\frac{{\left(3\right)}^{2}}{2}\mathrm{log}|\mathrm{t}+\sqrt{{\mathrm{t}}^{2}-{\left(3\right)}^{2}}|+\mathrm{C}\\ \mathrm{Putting}\text{t}=\mathrm{x}-4,\text{\hspace{0.17em}}\mathrm{we}\text{\hspace{0.17em}}\mathrm{get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{x}-4\right)}{2}\sqrt{{\left(\mathrm{x}-4\right)}^{2}-{\left(3\right)}^{2}}-\frac{9}{2}\mathrm{log}|\left(\mathrm{x}-4\right)+\sqrt{{\left(\mathrm{x}-4\right)}^{2}-{\left(3\right)}^{2}}|+\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{x}-4\right)}{2}\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}-\frac{9}{2}\mathrm{log}|\left(\mathrm{x}-4\right)+\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+7}|+\mathrm{C}\\ \mathrm{Hence},\text{the correct option is D.}\end{array}$