NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.4) Exercise 7.4

Mathematics is a high-scoring subject. The students of Class 12 should focus on practising Mathematics topics regularly. Studying it is essential for boosting the reasoning ability of students. Solving mathematics problems regularly is also helpful in approaching life problems in a better way. Mathematical fundamentals are used in other fields such as Economics, Physics, Accountancy, Social Science, etc.

It is important to access online resources to prepare well for the Mathematics exam. The NCERT Solutions for all the exercises of the Class 12 chapter named Integrals are available on the Extramarks’ website and mobile application in PDF format. All the questions asked in Exercise 7.4 of Chapter 7 can be solved with the help of the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.4. Students can learn to give accurate solutions in the board exam of Mathematics by using the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.4.

Class 12 is an important step in a student’s career. Scoring well in the Class 12 board examinations is critical for students to move ahead in their careers. The future of students largely depends upon the marks scored in the Class 12 Board exams. All the subjects are crucial from the board exam’s perspective. Students are supposed to focus on learning all the subjects in Class 12. Mathematics is one of the core subjects in the Class 12 curriculum. The students need to keep revising the Mathematical concepts in order to prepare well for the board exam.

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There are exercise questions asked on the basis of Class 12 Mathematics chapters. The students are advised to keep practising the exercise questions for scoring higher marks in the board exam of Mathematics.

Students are encouraged to solve the sample papers and past years’ papers to enhance their ongoing preparation for the Mathematics exam. Practising the past years’ papers of Mathematics is beneficial for understanding the type of questions that can appear in the board exam of Mathematics.

It is crucial to manage time while solving questions in mathematics. To improve the time management skills of students for the mathematics exam, it is advisable to practice exercise questions as much as possible. All the exercises in Chapter 7 are crucial to scoring well in the mathematics exam.

Students must access the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.4 from the Extramarks’ website and mobile application. The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.4 are curated by in-house experts in Mathematics. The solutions are helpful in breaking down complex problems into simpler and smaller steps that can be easily understood. Students are advised to make a detailed list of all the important topics and prepare them from perspective of the exam. Students need to solve the important questions of a chapter. It is crucial that students access study materials from credible sources.

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NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.4) Exercise 7.4

Students of Class 12 at the Central Board of Secondary Education (CBSE) can download the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.4 from the Extramarks’ website and mobile application in PDF format. All the solutions of Class 12th Math Exercise 7.4 are available in the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.4. The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.4 are given in an easy and understandable format.

Solving problems is beneficial for improving the analytical and reasoning skills of students. Students of Class 12 CBSE Board need to keep practising the exercise problems in order to increase their level of preparation for the Mathematics exam.

The topic of Chapter 7 in the Class 12 Mathematics textbook is Integrals. This topic in Class 12 is very important for preparing students for the broad concepts of Integrals. Students can download the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.4 from the Extramarks’ website and mobile application in PDF format. The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.4 are crucial for enhancing the learning process of Class 12 students. It is important to read the topics and subtopics of Chapter 7 from time to time. Students are advised to access the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.4 to enhance their understanding of the chapter Integrals.

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Learn More about Integrals and Integration

Integration is a crucial component of Calculus topics. The students are supposed to learn the Integrals concepts in detail. All the important formulas, theorems, definitions, graphs, etc. should be regularly practised. Students must revise all the topics of Integrals from time to time.

Questions 1 to 23 in Exercise 7.4 Class 12th Maths require students to integrate the given functions. The solutions of 1 to 23 questions can be practised with the help of the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.4. To integrate a given function, students must learn the definitions provided in the Integrals chapter and practice the application of formulas again and again.

Learn More about Integrals and Integration

The chapter Integrals is one of the crucial chapters in the Class 12 Mathematics syllabus. Students must go through each of the topics and subtopics in Chapter 7 thoroughly and understand their concepts. Students of Class 12 are required to learn about two types of Integrals. The Definite Integrals and Indefinite Integrals.

Definite Integrals are given by the area under a curve that consists of two fixed limits. An Indefinite Integral is an Integral without having limits. The Class 12 students of the Central Board of Secondary Education (CBSE) Board are advised to access the CBSE Revision Notes for revising the topics of the Integrals chapter.

The applications of integrals in the subject of Physics are:

Learning the application of Integrals is necessary. There are many applications of Integrals concepts in various fields of study. The concept of Integrals is used regularly in Physics. The knowledge of Integrals can be beneficial for learning Physics efficiently. Definite Integrals are used in Physics to determine the mass of an object. The topic of integration is also used in distance, velocity, and acceleration topics of physics.

NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.4) Exercise 7.4

There are 25 questions given in 12 Maths 7.4. Students can learn to solve each of the 25 questions with the help of the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.4. The Class 12 students of the Central Board of Secondary Education (CBSE) Board can download the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.4 from the Extramarks’ website and mobile application in PDF format. All the solutions given in the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.4 are available in a simple and clear format.

The NCERT Solutions are available for all the Classes of the CBSE Board on Extramarks.

Students of all the classes can access the NCERT solutions from Extramarks’ website and mobile application in PDF format.

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Importance of Integrals Chapter:

The concepts of Integrals are applied in various fields like Mathematics, Science, Engineering, etc. Students can learn to understand the application of Integrals by thoroughly reviewing the concepts and solving problems related to Integrals.

Definite Integrals can be used to solve problems in the fields of economics, finance, and probability. All the concepts of Integrals need to be consistently revised, and students must solve the problems from time to time. The Integration process consists of finding Integrals for functions. The Integrals concepts are used in Probability to find the probability of some random variable falling within a range.

The Integrals topics are used to find the area of a curve in a graph. Integrals are one of the essential topics of Calculus. It is important to learn Integrals topics in Class 12 efficiently in order to learn broad concepts of Calculus in the future.

Download Ex 7.4 Class 12 Maths NCERT Solutions PDF from Extramarks

Students of Class 12 can download the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.4 from the Extramarks’ website and mobile application in PDF format. The study materials provided by Extramarks are very trustworthy and easy to understand. Extramarks is dedicated to make the learning process easier for the students. The live classes for all the subjects are conducted on the Extramarks website and mobile application. These live classes are very interactive, and students can easily get solutions to their problems from expert teachers in all the subjects. The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.4 are available on the Extramarks’ website and mobile application in PDF format.

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Benefits of Downloading NCERT Maths Ex 7.4 Class 12 Solutions From Extramarks

The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.4 are helpful in boosting the confidence of students. Class 12 students of the Central Board of Secondary Education (CBSE) Board are advised to get an overview of the CBSE Mathematics Syllabus. It is advisable to prepare a unique study plan for learning mathematics topics. The syllabus is helpful in devising a strategy for learning chapters.

Students can learn to manage time effectively by practising Exercise 7.4 with the help of the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.4. The questions of Exercise 7.4 are based on the topic of Integrals of Some Particular Functions. Students need to read the integrals of some particular functions topic and understand its meaning.

All the questions in Exercise 7.4 are important in the context of the board exam. Students must practice the questions in Exercise 7.4 consistently. Some students may find it difficult to solve the questions in Exercise 7.4. Thus, students can make use of the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.4 to get detailed solutions for Exercise 7.4.

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Q.1 $Integrate the function given below . \frac{3 x^{2}}{x^{6} + 1}$

Ans

$\begin{array}{l} Let t = x^{3} \Rightarrow \frac{dt}{dx} = 3 x^{2} \\ \Rightarrow dx = \frac{dt}{3 x^{2}} \\ ∴ \int \frac{3 x^{2}}{x^{6} + 1} dx = \int \frac{3 x^{2}}{{(x^{3})}^{2} + 1} dx \\ = \int \frac{3 x^{2}}{t^{2} + 1} . \frac{dt}{3 x^{2}} \\ = \int \frac{1}{t^{2} + 1} dt \\ = \tan^{- 1} t + C \\ = \tan^{- 1} (x^{3}) + C \end{array}$

Q.2

$Integrate the function given below . \frac{1}{\sqrt{1 + 4 x^{2}}}$

Ans

$\begin{array}{l} We have, \frac{1}{\sqrt{1 + 4 x^{2}}} = \frac{1}{\sqrt{1 + {(2 x)}^{2}}} \\ Let t = 2 x \Rightarrow \frac{dt}{dx} = 2 \\ \Rightarrow dx = \frac{dt}{2} \\ ∴ \int \frac{1}{\sqrt{1 + 4 x^{2}}} dx = \int \frac{1}{\sqrt{1 + {(2 x)}^{2}}} dx \\ = \int \frac{1}{\sqrt{1 + t^{2}}} . \frac{dt}{2} \\ = \frac{1}{2} \int \frac{1}{\sqrt{1 + t^{2}}} dt \\ = \frac{1}{2} \log | t + \sqrt{t^{2} + 1} | + C \\ = \frac{1}{2} \log | 2 x + \sqrt{{(2 x)}^{2} + 1} | + C \\ = \frac{1}{2} \log | 2 x + \sqrt{4 x^{2} + 1} | + C \end{array}$

Q.3

$Integrate the function given below . \frac{1}{\sqrt{{(2 - x)}^{2} + 1}}$

Ans

$\begin{array}{l} We have, \frac{1}{\sqrt{{(2 - x)}^{2} + 1}} \\ Let t = 2 - x \Rightarrow \frac{dt}{dx} = - 1 \\ \Rightarrow dx = - dt \\ ∴ \int \frac{1}{\sqrt{{(2 - x)}^{2} + 1}} dx = \int \frac{1}{\sqrt{t^{2} + 1}} (- dt) \\ = - \int \frac{1}{\sqrt{1 + t^{2}}} . dt \\ = - \log | t + \sqrt{t^{2} + 1} | + C [\begin{array}{l} ∵ \int \frac{1}{\sqrt{x^{2} + a^{2}}} dx \\ = \log | x + \sqrt{x^{2} + a^{2}} | \end{array}] \\ = - \log | (2 - x) + \sqrt{{(2 - x)}^{2} + 1} | + C \\ = \log | \frac{1}{(2 - x) + \sqrt{{(2 - x)}^{2} + 1}} | + C [∵ - logx = \log (\frac{1}{x})] \\ = \log | \frac{1}{(2 - x) + \sqrt{4 - 4 x + x^{2} + 1}} | + C \\ = \log | \frac{1}{(2 - x) + \sqrt{x^{2} - 4 x + 5}} | + C \end{array}$

Q.4

$Integrate the function given below . \frac{1}{\sqrt{9 - 25 x^{2}}}$

Ans

$\begin{array}{l} We have, \frac{1}{\sqrt{9 - 25 x^{2}}} = \frac{1}{\sqrt{9 - {(5 x)}^{2}}} \\ Let t = 5 x \Rightarrow \frac{dt}{dx} = 5 \\ \Rightarrow dx = \frac{1}{5} dt \\ ∴ \int \frac{1}{\sqrt{9 - 25 x^{2}}} dx = \frac{1}{5} \int \frac{1}{\sqrt{9 - {(5 x)}^{2}}} dt \\ = \frac{1}{5} \int \frac{1}{\sqrt{3^{2} - t^{2}}} . dt \\ = \frac{1}{5} \sin^{- 1} (\frac{t}{3}) + C \\ = \frac{1}{5} \sin^{- 1} (\frac{5 x}{3}) + C \end{array}$

Q.5

$Integrate the function \frac{3 x}{1 + 2 x^{4}}$

Ans

$\begin{array}{l} We have, \frac{3 x}{1 + 2 x^{4}} = \frac{3 x}{1 + {(\sqrt{2} x^{2})}^{2}} \\ Let t = \sqrt{2} x^{2} \Rightarrow \frac{dt}{dx} = 2 \sqrt{2} x \\ \Rightarrow dx = \frac{dt}{2 \sqrt{2} x} \\ ∴ \int \frac{3 x}{1 + 2 x^{4}} dx = \int \frac{3 x}{1 + {(\sqrt{2} x^{2})}^{2}} dx \\ = \int \frac{3 x}{1 + t^{2}} . \frac{dt}{2 \sqrt{2} x} \\ = \frac{3}{2 \sqrt{2}} \int \frac{1}{1 + t^{2}} dt \\ = \frac{3}{2 \sqrt{2}} \tan^{- 1} t + C = \frac{3}{2 \sqrt{2}} \tan^{- 1} (\sqrt{2} x^{2}) + C \end{array}$

Q.6

$Integrate the function \frac{x^{2}}{1 - x^{6}}$

Ans

$\begin{array}{l} We have, \frac{x^{2}}{1 - x^{6}} = \frac{x^{2}}{1 - {(x^{3})}^{2}} \\ Let t = x^{3} \Rightarrow \frac{dt}{dx} = 3 x^{2} \\ \Rightarrow dx = \frac{dt}{3 x^{2}} \\ ∴ \int \frac{x^{2}}{1 - x^{6}} dx = \int \frac{x^{2}}{1 - {(x^{3})}^{2}} dx \\ = \int \frac{x^{2}}{1 - t^{2}} . \frac{dt}{3 x^{2}} \\ = \frac{1}{3} \int \frac{1}{1 - t^{2}} dt \\ = \frac{1}{3} (\frac{1}{2} \log | \frac{1 + t}{1 - t} |) + C \\ = \frac{1}{6} \log | \frac{1 + x^{3}}{1 - x^{3}} | + C \end{array}$

Q.7

$Integrate the function \frac{x - 1}{\sqrt{x^{2} - 1}}$

Ans

$\begin{array}{l} We have, \frac{x - 1}{\sqrt{x^{2} - 1}} = \frac{x - 1}{\sqrt{x^{2} - 1}} \\ Let t = x^{2} - 1 \Rightarrow \frac{dt}{dx} = 2 x \\ \Rightarrow dx = \frac{dt}{2 x} \\ ∴ \int \frac{x - 1}{\sqrt{x^{2} - 1}} dx = \int \frac{x}{\sqrt{x^{2} - 1}} dx - \int \frac{1}{\sqrt{x^{2} - 1}} dx \\ = \int \frac{x}{\sqrt{t}} . \frac{dt}{2 x} - \int \frac{1}{\sqrt{x^{2} - 1}} dx \\ = \frac{1}{2} \int t^{- \frac{1}{2}} . dt - \log | x + \sqrt{x^{2} - 1} | + C \\ = \frac{1}{2} (2 \sqrt{t}) - \log | x + \sqrt{x^{2} - 1} | + C \\ = \sqrt{t} - \log | x + \sqrt{x^{2} - 1} | + C \\ = \sqrt{x^{2} - 1} - \log | x + \sqrt{x^{2} - 1} | + C \end{array}$

Q.8

$Integrate the function \frac{x^{2}}{\sqrt{x^{6} + a^{6}}}$

Ans

$\begin{array}{l} We have, \frac{x^{2}}{\sqrt{x^{6} + a^{6}}} = \frac{x^{2}}{\sqrt{a^{6} + {(x^{3})}^{2}}} \\ Let t = x^{3} \Rightarrow \frac{dt}{dx} = 3 x^{2} \\ \Rightarrow dx = \frac{dt}{3 x^{2}} \\ ∴ \int \frac{x^{2}}{\sqrt{x^{6} + a^{6}}} dx = \int \frac{x^{2}}{\sqrt{{(a^{3})}^{2} + {(x^{3})}^{2}}} dx \\ = \int \frac{x^{2}}{\sqrt{{(a^{3})}^{2} + t^{2}}} . \frac{dt}{3 x^{2}} \\ = \frac{1}{3} \int \frac{1}{\sqrt{{(a^{3})}^{2} + t^{2}}} dt \\ = \frac{1}{3} \log | t + \sqrt{t^{2} + {(a^{3})}^{2}} | + C \\ = \frac{1}{3} \log | x^{3} + \sqrt{{(x^{3})}^{2} + {(a^{3})}^{2}} | + C \\ = \frac{1}{3} \log | x^{3} + \sqrt{x^{6} + a^{6}} | + C \end{array}$

Q.9

$Integrate the function \frac{\sec^{2} x}{\sqrt{\tan^{2} x + 4}}$

Ans

$\begin{array}{l} We have, \frac{\sec^{2} x}{\sqrt{\tan^{2} x + 4}} = \frac{\sec^{2} x}{\sqrt{\tan^{2} x + 2^{2}}} \\ Let t = tanx \Rightarrow \frac{dt}{dx} = \sec^{2} x \\ \Rightarrow dx = \frac{dt}{\sec^{2} x} \\ ∴ \int \frac{\sec^{2} x}{\sqrt{\tan^{2} x + 4}} dx = \int \frac{\sec^{2} x}{\sqrt{\tan^{2} x + 2^{2}}} dx \\ = \int \frac{\sec^{2} x}{\sqrt{t^{2} + 2^{2}}} . \frac{dt}{\sec^{2} x} \\ = \int \frac{1}{\sqrt{t^{2} + 2^{2}}} dt \\ = \log | t + \sqrt{t^{2} + {(2)}^{2}} | + C \\ = \log | tanx + \sqrt{\tan^{2} x + 4} | + C \end{array}$

Q.10

$Integrate the function \frac{1}{\sqrt{x^{2} + 2 x + 2}}$

Ans

$\begin{array}{l} We have, \\ \frac{1}{\sqrt{x^{2} + 2 x + 2}} = \frac{1}{\sqrt{{(x + 1)}^{2} + 1}} \\ Let t = x + 1 \Rightarrow \frac{dt}{dx} = 1 \\ \Rightarrow dx = dt \\ ∴ \int \frac{1}{\sqrt{x^{2} + 2 x + 2}} dx = \int \frac{1}{\sqrt{{(x + 1)}^{2} + 1}} dx \\ = \int \frac{1}{\sqrt{t^{2} + 1^{2}}} . dt \\ = \log | t + \sqrt{t^{2} + {(1)}^{2}} | + C \\ = \log | (x + 1) + \sqrt{{(x + 1)}^{2} + 1} | + C \\ = \log | (x + 1) + \sqrt{x^{2} + 2 x + 2} | + C \end{array}$

Q.11

$Integrate the function \frac{1}{9 x^{2} + 6 x + 5}$

Ans

$\begin{array}{l} We have, \frac{1}{9 x^{2} + 6 x + 5} = \frac{1}{{(3 x + 1)}^{2} + 2^{2}} \\ Let t = 3 x + 1 \Rightarrow \frac{dt}{dx} = 3 \\ \Rightarrow dx = \frac{dt}{3} \\ ∴ \int \frac{1}{9 x^{2} + 6 x + 5} dx = \int \frac{1}{{(3 x + 1)}^{2} + 2^{2}} dx \\ = \int \frac{1}{t^{2} + 2^{2}} . \frac{dt}{3} \\ = \frac{1}{3} \int \frac{1}{t^{2} + 2^{2}} dt \\ = \frac{1}{3} (\frac{1}{2} \tan^{- 1} \frac{t}{2}) + C \\ = \frac{1}{6} \tan^{- 1} \frac{(3 x + 1)}{2} + C \end{array}$

Q.12

$Integrate the function \frac{1}{\sqrt{7 - 6 x - x^{2}}}$

Ans

$\begin{array}{l} We have, \frac{1}{\sqrt{7 - 6 x - x^{2}}} = \frac{1}{\sqrt{16 - {(x + 3)}^{2}}} \\ Let t = x + 3 \Rightarrow \frac{dt}{dx} = 1 \\ \Rightarrow dx = dt \\ ∴ \int \frac{1}{\sqrt{7 - 6 x - x^{2}}} dx = \int \frac{1}{\sqrt{16 - {(x + 3)}^{2}}} dx \\ = \int \frac{1}{\sqrt{4^{2} - t^{2}}} dt \\ = \sin^{- 1} (\frac{t}{4}) + C \\ = \sin^{- 1} (\frac{x + 3}{4}) + C \end{array}$

Q.13

$Integrate the function \frac{1}{\sqrt{(x - 1) (x - 2)}}$

Ans

$\begin{array}{l} We have, \frac{1}{\sqrt{(x - 1) (x - 2)}} = \frac{1}{\sqrt{x^{2} - 3 x + 2}} \\ = \frac{1}{\sqrt{{(x - \frac{3}{2})}^{2} - \frac{1}{4}}} \\ Let t = x - \frac{3}{2} \Rightarrow \frac{dt}{dx} = 1 \\ \Rightarrow dx = dt \\ ∴ \int \frac{1}{\sqrt{(x - 1) (x - 2)}} dx = \int \frac{1}{\sqrt{{(x - \frac{3}{2})}^{2} - \frac{1}{4}}} dx \\ = \int \frac{1}{\sqrt{t^{2} - {(\frac{1}{2})}^{2}}} dt \\ = \log | t + \sqrt{t^{2} - {(\frac{1}{2})}^{2}} | + C \\ = \log | (x - \frac{3}{2}) + \sqrt{{(x - \frac{3}{2})}^{2} - {(\frac{1}{2})}^{2}} | + C \\ = \log | (x - \frac{3}{2}) + \sqrt{x^{2} - 3 x + 2} | + C \end{array}$

Q.14

$Integrate the function \frac{1}{\sqrt{8 + 3 x - x^{2}}}$

Ans

$\begin{array}{l} We have, \frac{1}{\sqrt{8 + 3 x - x^{2}}} = \frac{1}{\sqrt{8 - (x^{2} - 3 x + \frac{9}{4} - \frac{9}{4})}} \\ = \frac{1}{\sqrt{8 + \frac{9}{4} - {(x - \frac{3}{2})}^{2}}} \\ = \frac{1}{\sqrt{\frac{41}{4} - {(x - \frac{3}{2})}^{2}}} \\ Let t = x - \frac{3}{2} \Rightarrow \frac{dt}{dx} = 1 \\ \Rightarrow dx = dt \\ ∴ \int \frac{1}{\sqrt{8 + 3 x - x^{2}}} dx = \int \frac{1}{\sqrt{\frac{41}{4} - {(x - \frac{3}{2})}^{2}}} dx \\ = \int \frac{1}{\sqrt{{(\frac{\sqrt{41}}{2})}^{2} - t^{2}}} dt \\ = \sin^{- 1} (\frac{t}{\frac{\sqrt{41}}{2}}) + C \\ = \sin^{- 1} (\frac{2 t}{\sqrt{41}}) + C \\ = \sin^{- 1} {\frac{2 (x - \frac{3}{2})}{\sqrt{41}}} + C \\ = \sin^{- 1} {\frac{(2 x - 3)}{\sqrt{41}}} + C \end{array}$

Q.15

$Integrate the function \frac{1}{\sqrt{(x - a) (x - b)}}$

Ans

$\begin{array}{l} We have, \frac{1}{\sqrt{(x - a) (x - b)}} = \frac{1}{\sqrt{x^{2} - (a + b) x + ab}} \\ = \frac{1}{\sqrt{{(x - \frac{a + b}{2})}^{2} - {(\frac{a + b}{2})}^{2}}} \\ Let t = x - \frac{a + b}{2} \Rightarrow \frac{dt}{dx} = 1 \\ \Rightarrow dx = dt \\ ∴ \int \frac{1}{\sqrt{(x - a) (x - b)}} dx = \int \frac{1}{\sqrt{{(x - \frac{a + b}{2})}^{2} - {(\frac{a + b}{2})}^{2}}} dx \\ = \int \frac{1}{\sqrt{t^{2} - {(\frac{a + b}{2})}^{2}}} dt \\ = \int \frac{1}{\sqrt{t^{2} - {(\frac{a + b}{2})}^{2}}} dt \\ = \log | t + \sqrt{t^{2} - {(\frac{a + b}{2})}^{2}} | + C \\ = \log | (x - \frac{a + b}{2}) + \sqrt{{(x - \frac{a + b}{2})}^{2} - {(\frac{a + b}{2})}^{2}} | + C \\ = \log | (x - \frac{a + b}{2}) + \sqrt{x^{2} - (a + b) x + ab} | + C \end{array}$

Q.16

$Integrate the function \frac{4 x + 1}{\sqrt{2 x^{2} + x - 3}}$

Ans

$\begin{array}{l} We have, \frac{4 x + 1}{\sqrt{2 x^{2} + x - 3}} \\ Here, 4 x + 1 = A \frac{d}{dx} (2 x^{2} + x - 3) + B \\ = A (4 x + 1) + B \\ Equating the coefficients of x and the contant terms from \\ both sides, we get \\ 4 = 4 A and A + B = 1 \Rightarrow A = 1 and B = 1 - 1 = 0 \\ Now, let t = 2 x^{2} + x - 3 \Rightarrow \frac{dt}{dx} = 4 x + 1 \\ \Rightarrow dx = \frac{dt}{4 x + 1} \\ ∴ \int \frac{4 x + 1}{\sqrt{2 x^{2} + x - 3}} dx = \int \frac{4 x + 1}{\sqrt{2 x^{2} + x - 3}} . \frac{dt}{4 x + 1} \\ = \int \frac{1}{\sqrt{t}} . dt \\ = 2 \sqrt{t} + C \\ = 2 \sqrt{2 x^{2} + x - 3} + C \end{array}$

Q.17

$Integrate the function \frac{x + 2}{\sqrt{x^{2} - 1}}$

Ans

$\begin{array}{l} We have, \frac{x + 2}{\sqrt{x^{2} - 1}} \\ Here, x + 2 = A \frac{d}{dx} (x^{2} - 1) + B \\ = A (2 x) + B \\ Equating the coefficients of x and the contant terms from \\ both sides, we get \\ 1 = 2 A and B = 2 \Rightarrow A = \frac{1}{2} and B = 2 \\ Therefore, x + 2 = \frac{1}{2} (2 x) + 2 \\ ∴ \int \frac{x + 2}{\sqrt{x^{2} - 1}} dx = \int \frac{\frac{1}{2} (2 x) + 2}{\sqrt{x^{2} - 1}} . dx \\ = \int \frac{\frac{1}{2} (2 x)}{\sqrt{x^{2} - 1}} dx + \int \frac{2}{\sqrt{x^{2} - 1}} dx \\ Now, let t = x^{2} - 1 \Rightarrow \frac{dt}{dx} = 2 x \\ \Rightarrow dx = \frac{dt}{2 x} \\ then \\ \int \frac{x + 2}{\sqrt{x^{2} - 1}} dx = \frac{1}{2} \int \frac{(2 x)}{\sqrt{t}} . \frac{dt}{2 x} + \int \frac{2}{\sqrt{x^{2} - 1}} dx \\ = \int \frac{1}{\sqrt{t}} . dt + 2 \log | x + \sqrt{x^{2} - 1} | \\ = \sqrt{t} + 2 \log | x + \sqrt{x^{2} - 1} | + C \\ = \sqrt{x^{2} - 1} + 2 \log | x + \sqrt{x^{2} - 1} | + C \end{array}$

Q.18

$Integrate the function \frac{5 x - 2}{1 + 2 x + 3 x^{2}}$

Ans

$\begin{array}{l} We have, \frac{5 x - 2}{1 + 2 x + 3 x^{2}} \\ Here, 5 x - 2 = A \frac{d}{dx} (1 + 2 x + 3 x^{2}) + B \\ = A (6 x + 2) + B \\ Equating the coefficients of x and the contant terms from \\ both sides, we get \\ 5 = 6 A and 2A + B = - 2 \\ \Rightarrow A = \frac{5}{6} and B = - 2 - 2 (\frac{5}{6}) = - 2 - \frac{5}{3} = - \frac{11}{3} \\ ∴ 5 x - 2 = \frac{5}{6} (6 x + 2) - \frac{11}{3} \\ ∴ \int \frac{5 x - 2}{1 + 2 x + 3 x^{2}} dx = \int \frac{\frac{5}{6} (6 x + 2) - \frac{11}{3}}{\sqrt{2 x^{2} + x - 3}} dx \\ = \frac{5}{6} \int \frac{(6 x + 2)}{\sqrt{1 + 2 x + 3 x^{2}}} dx - \frac{11}{3} \int \frac{1}{\sqrt{1 + 2 x + 3 x^{2}}} dx \\ Now, let t = 1 + 2 x + 3 x^{2} \Rightarrow \frac{dt}{dx} = 6 x + 2 \\ \Rightarrow dx = \frac{dt}{6 x + 2}, then \\ \int \frac{5 x - 2}{1 + 2 x + 3 x^{2}} dx = \frac{5}{6} \int \frac{(6 x + 2)}{1 + 2 x + 3 x^{2}} . \frac{dt}{6 x + 2} - \frac{11}{3} \int \frac{1}{1 + 2 x + 3 x^{2}} . dx \\ = \frac{5}{6} \int \frac{1}{t} . dt - \frac{11}{3} \int \frac{1}{3 {{(x + \frac{1}{3})}^{2} + {(\frac{\sqrt{2}}{3})}^{2}}} . dx \\ = \frac{5}{6} \log | t | - \frac{11}{9} \int \frac{1}{{(x + \frac{1}{3})}^{2} + {(\frac{\sqrt{2}}{3})}^{2}} . dx \\ = \frac{5}{6} \log | 1 + 2 x + 3 x^{2} | - \frac{11}{9} \times \frac{1}{(\frac{\sqrt{2}}{3})} \tan^{- 1} (\frac{x + \frac{1}{3}}{\frac{\sqrt{2}}{3}}) + C \\ = \frac{5}{6} \log | 1 + 2 x + 3 x^{2} | - \frac{11}{3 \sqrt{2}} \tan^{- 1} (\frac{3 x + 1}{\sqrt{2}}) + C \end{array}$

Q.19

$Integrate the function \frac{6 x + 7}{\sqrt{(x - 5) (x - 4)}}$

Ans

$\begin{array}{l} We have, \frac{6 x + 7}{\sqrt{(x - 5) (x - 4)}} = \frac{6 x + 7}{\sqrt{x^{2} - 9 x + 20}} \\ Here, 6 x + 7 = A \frac{d}{dx} (x^{2} - 9 x + 20) + B \\ 6 x + 7 = A (2 x - 9) + B \\ Equating the coefficients of x and the contant terms from \\ both sides, we get \\ 6 = 2 A and - 9A + B = 7 \\ \Rightarrow A = \frac{6}{2} = 3 and B = 7 + 9 (3) = 34 \\ ∴ 6 x + 7 = 3 (2 x - 9) + 34 \\ ∴ \int \frac{6 x + 7}{\sqrt{(x - 5) (x - 4)}} dx = \int \frac{6 x + 7}{\sqrt{x^{2} - 9 x + 20}} . dx \\ = \int \frac{3 (2 x - 9) + 34}{\sqrt{x^{2} - 9 x + 20}} . dx \\ = 3 \int \frac{(2 x - 9)}{\sqrt{x^{2} - 9 x + 20}} dx + \int \frac{34}{\sqrt{x^{2} - 9 x + 20}} dx \\ Now, let t = x^{2} - 9 x + 20 \Rightarrow \frac{dt}{dx} = 2 x - 9 \\ \Rightarrow dx = \frac{dt}{2 x - 9}, then \\ \int \frac{6 x + 7}{\sqrt{(x - 5) (x - 4)}} dx = 3 \int \frac{(2 x - 9)}{\sqrt{t}} . \frac{dt}{2 x - 9} + 34 \int \frac{1}{\sqrt{x^{2} - 9 x + 20}} . dx \\ = 3 \int \frac{1}{\sqrt{t}} . dt + 34 \int \frac{1}{\sqrt{\{{(x - \frac{9}{2})}^{2} - {(\frac{1}{2})}^{2}\}}} . dx \\ = 3 \times 2 \sqrt{t} + 34 \log |(x - \frac{9}{2}) + \sqrt{\{{(x - \frac{9}{2})}^{2} - {(\frac{1}{2})}^{2}\}}| + C \\ = 6 \sqrt{x^{2} - 9 x + 20} + 34 \log |(x - \frac{9}{2}) + \sqrt{x^{2} - 9 x + 20}| + C \end{array}$

Q.20

$Integrate the function \frac{x + 2}{\sqrt{4 x - x^{2}}}$

Ans

$\begin{array}{l} We have, \frac{x + 2}{\sqrt{4 x - x^{2}}} \\ Here, x + 2 = A \frac{d}{dx} (4 x - x^{2}) + B \\ x + 2 = A (4 - 2 x) + B \\ Equating the coefficients of x and the contant terms from \\ both sides, we get \\ 1 = - 2 A and 4A + B = 2 \\ \Rightarrow A = - \frac{1}{2} and B = 2 - 4 (- \frac{1}{2}) = 4 \\ ∴ x + 2 = - \frac{1}{2} (4 - 2 x) + 4 \\ ∴ \int \frac{x + 2}{\sqrt{4 x - x^{2}}} dx = \int \frac{- \frac{1}{2} (4 - 2 x) + 4}{\sqrt{4 x - x^{2}}} dx \\ = - \frac{1}{2} \int \frac{(4 - 2 x)}{\sqrt{4 x - x^{2}}} dx + \int \frac{4}{\sqrt{4 x - x^{2}}} dx \\ = - \frac{1}{2} \int \frac{(4 - 2 x)}{\sqrt{4 x - x^{2}}} dx + \int \frac{4}{\sqrt{4 - (4 - 4 x + x^{2})}} . dx \\ = - \frac{1}{2} \int \frac{(4 - 2 x)}{\sqrt{4 x - x^{2}}} dx + \int \frac{4}{\sqrt{2^{2} - {(2 - x)}^{2}}} dx \\ Now, let t = 4 x - x^{2} \Rightarrow \frac{dt}{dx} = 4 - 2 x \\ \Rightarrow dx = \frac{dt}{4 - 2 x}, then \\ \int \frac{x + 2}{\sqrt{4 x - x^{2}}} dx = - \frac{1}{2} \int \frac{(4 - 2 x)}{\sqrt{t}} \frac{dt}{4 - 2 x} + \int \frac{4}{\sqrt{2^{2} - {(2 - x)}^{2}}} dx \\ = - \frac{1}{2} \times 2 \sqrt{t} - 4 \sin^{- 1} (\frac{2 - x}{2}) + C \\ = - \sqrt{4 x - x^{2}} - 4 \sin^{- 1} (\frac{2 - x}{2}) + C \end{array}$

Q.21

$Integrate the function \frac{x + 2}{\sqrt{x^{2} + 2 x + 3}}$

Ans

$\begin{array}{l} We have, \frac{x + 2}{\sqrt{x^{2} + 2 x + 3}} \\ \int \frac{x + 2}{\sqrt{x^{2} + 2 x + 3}} dx = \frac{1}{2} \int \frac{2 (x + 2)}{\sqrt{x^{2} + 2 x + 3}} dx \\ = \frac{1}{2} \int \frac{2 x + 4}{\sqrt{x^{2} + 2 x + 3}} dx \\ = \frac{1}{2} \int \frac{2 x + 2}{\sqrt{x^{2} + 2 x + 3}} dx + \frac{1}{2} \int \frac{2}{\sqrt{x^{2} + 2 x + 3}} dx \\ Let t = x^{2} + 2 x + 3 \Rightarrow \frac{dt}{dx} = 2 x + 2 \\ \Rightarrow dx = \frac{dt}{2 x + 2}, then \\ \int \frac{x + 2}{\sqrt{x^{2} + 2 x + 3}} dx = \frac{1}{2} \int \frac{2 x + 2}{\sqrt{t}} \frac{dt}{2 x + 2} + \frac{1}{2} \int \frac{2}{\sqrt{x^{2} + 2 x + 3}} dx \\ = \frac{1}{2} \int \frac{1}{\sqrt{t}} dt + \int \frac{1}{\sqrt{{(x + 1)}^{2} + {(\sqrt{2})}^{2}}} dx \\ = \frac{1}{2} . 2 \sqrt{t} + \log | (x + 1) + \sqrt{{(x + 1)}^{2} + {(\sqrt{2})}^{2}} | + C \\ = \sqrt{x^{2} + 2 x + 3} + \log | (x + 1) + \sqrt{x^{2} + 2 x + 3} | + C \end{array}$

Q.22

$Integrate the function \frac{x + 3}{x^{2} - 2 x - 5}$

Ans

$\begin{array}{l} We have, \frac{x + 3}{x^{2} - 2 x - 5} \\ Here, x + 3 = A \frac{d}{dx} (x^{2} - 2 x - 5) + B \\ x + 3 = A (2 x - 2) + B \\ Equating the coefficients of x and the contant terms from \\ both sides, we get \\ 1 = 2 A and - 2A + B = 3 \\ \Rightarrow A = \frac{1}{2} and B = 3 + 2 (\frac{1}{2}) = 4 \\ ∴ x + 3 = \frac{1}{2} (2 x - 2) + 4 \\ \Rightarrow \int \frac{x + 3}{x^{2} - 2 x - 5} dx = \int \frac{\frac{1}{2} (2 x - 2) + 4}{x^{2} - 2 x - 5} dx \\ = \frac{1}{2} \int \frac{(2 x - 2)}{x^{2} - 2 x - 5} dx + 4 \int \frac{1}{x^{2} - 2 x - 5} dx \\ Let t = x^{2} - 2 x - 5 \Rightarrow \frac{dt}{dx} = 2 x - 2 \\ \Rightarrow dx = \frac{dt}{2 x - 2}, then \\ \Rightarrow \int \frac{x + 3}{x^{2} - 2 x - 5} dx = \frac{1}{2} \int \frac{(2 x - 2)}{t} \frac{dt}{2 x - 2} + 4 \int \frac{1}{{(x - 1)}^{2} - 6} dx \\ = \frac{1}{2} \int \frac{1}{t} dt + 4 \int \frac{1}{{(x - 1)}^{2} - {(\sqrt{6})}^{2}} dx \\ = \frac{1}{2} \log | t | + 4 \times \frac{1}{2 \sqrt{6}} \log (\frac{x - 1 - \sqrt{6}}{x - 1 + \sqrt{6}}) + C \\ = \frac{1}{2} \log | x^{2} - 2 x - 5 | + \frac{2}{\sqrt{6}} \log (\frac{x - 1 - \sqrt{6}}{x - 1 + \sqrt{6}}) + C \end{array}$

Q.23

$Integrate the function \frac{5 x + 3}{\sqrt{x^{2} + 4 x + 10}}$

Ans

$\begin{array}{l} We have, \frac{5 x + 3}{\sqrt{x^{2} + 4 x + 10}} \\ Here, 5 x + 3 = A \frac{d}{dx} (x^{2} + 4 x + 10) + B \\ 5 x + 3 = A (2 x + 4) + B \\ Equating the coefficients of x and the contant terms from \\ both sides, we get \\ 5 = 2 A and 4A + B = 3 \\ \Rightarrow A = \frac{5}{2} and B = 3 - 4 (\frac{5}{2}) = - 7 \\ ∴ 5 x + 3 = \frac{5}{2} (2 x + 4) - 7 \\ ∴ \int \frac{5 x + 3}{\sqrt{x^{2} + 4 x + 10}} dx = \int \frac{\frac{5}{2} (2 x + 4) - 7}{\sqrt{x^{2} + 4 x + 10}} dx \\ = \frac{5}{2} \int \frac{(2 x + 4)}{\sqrt{x^{2} + 4 x + 10}} dx - \int \frac{7}{\sqrt{x^{2} + 4 x + 10}} dx \\ = \frac{5}{2} \int \frac{(2 x + 4)}{\sqrt{x^{2} + 4 x + 10}} dx - \int \frac{7}{\sqrt{{(x + 2)}^{2} + 6}} dx \\ = \frac{5}{2} \int \frac{(2 x + 4)}{\sqrt{x^{2} + 4 x + 10}} dx - 7 \int \frac{1}{\sqrt{{(x + 2)}^{2} + {(\sqrt{6})}^{2}}} dx \\ Now, let t = x^{2} + 4 x + 10 \Rightarrow \frac{dt}{dx} = 2 x + 4 \\ \Rightarrow dx = \frac{dt}{2 x + 4}, then \\ \int \frac{5 x + 3}{\sqrt{x^{2} + 4 x + 10}} dx = \frac{5}{2} \int \frac{(2 x + 4)}{\sqrt{t}} \frac{dt}{2 x + 4} - 7 \int \frac{1}{\sqrt{{(x + 2)}^{2} + {(\sqrt{6})}^{2}}} dx \\ = \frac{5}{2} \times 2 \sqrt{t} - 7 \log | (x + 2) + \sqrt{{(x + 2)}^{2} + {(\sqrt{6})}^{2}} | + C \\ = 5 \sqrt{x^{2} + 4 x + 10} - 7 \log | (x + 2) + \sqrt{x^{2} + 4 x + 10} | + C \end{array}$

Q.24

$\begin{array}{l} Choose the correct answer \\ \int \frac{dx}{x^{2} + 2 x + 2} \\ (A) x \tan^{- 1} (x + 1) + C \\ (B) \tan^{- 1} (x + 1) + C \\ (C) (x + 1) \tan^{- 1} x + C \\ (D) \tan^{- 1} x + C \end{array}$

Ans

$\begin{array}{l} We have, \\ \int \frac{dx}{x^{2} + 2 x + 2} = \int \frac{dx}{{(x + 1)}^{2} + 1^{2}} \\ = \tan^{- 1} (\frac{x + 1}{1}) + C \\ = \tan^{- 1} (x + 1) + C \\ Hence, the correct option is B. \end{array}$

Q.25

$\begin{array}{l} Choose the correct answer \\ \int \frac{dx}{\sqrt{9 x - 4 x^{2}}} equals \\ (A) \frac{1}{9} \sin^{- 1} (\frac{9 x - 8}{8}) + C \\ (B) \frac{1}{2} \sin^{- 1} (\frac{8 x - 9}{9}) + C \\ (C) \frac{1}{3} \sin^{- 1} (\frac{9 x - 8}{8}) + C \\ (D) \frac{1}{2} \sin^{- 1} (\frac{9 x - 8}{9}) + C \end{array}$

Ans

$\begin{array}{l} \int \frac{dx}{\sqrt{9 x - 4 x^{2}}} = \int \frac{dx}{\sqrt{- 4 (x^{2} - \frac{9}{4} x)}} \\ = \int \frac{dx}{\sqrt{- 4 {x^{2} - \frac{9}{4} x + {(\frac{9}{8})}^{2} - {(\frac{9}{8})}^{2}}}} \\ = \frac{1}{2} \int \frac{dx}{\sqrt{{(\frac{9}{8})}^{2} - {(x - \frac{9}{8})}^{2}}} \\ = \frac{1}{2} \sin^{- 1} (\frac{x - \frac{9}{8}}{\frac{9}{8}}) + C \\ = \frac{1}{2} \sin^{- 1} (\frac{8 x - 9}{9}) + C \\ Hence, the correct option is B. \end{array}$

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NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.4) Exercise 7.4

NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.4) Exercise 7.4

NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.4) Exercise 7.4

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NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.4) Exercise 7.4

NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.4) Exercise 7.4

NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.4) Exercise 7.4

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2. How can the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.4 benefit Class 12 students?

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4. What are the benefits of accessing the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.4 from Extramarks?

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