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# Class 12 Maths NCERT solutions of Exercise 7.4-Integrals

Class 12 Maths NCERT solutions of Exercise 7.4 of chapter 7-Integrals are created by an expert teacher of maths. These NCERT solutions are the solutions to the unsolved questions of class 12 maths exercise 7.4 of chapter 7-Integrals of the NCERT textbook prescribed by the CBSE School board of India.NCERT solutions of exercise 7.4 will help in your preparation of the board exam of CBSE and in clearing the ways of methods of integration used in different kinds of functions. By studying these NCERT Solutions you will become capable to solve all problems related to the integration.

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In solving unsolved questions of exercise 7.4 of chapter 7-Integrals you have to understand solutions of algebraic expressions so you can revise the maths chapter of previous classes like quadratic equations. Click here to revise complete square of polynomial

## Class 12 Maths NCERT solutions of Chapter 7 Integrals

Exercise 7.1- Integrals

Exercise 7.2-Integral

Exercise 7.3-Integrals

Exercise 7.4 -Integral

## Class 12 Maths NCERT solutions of Exercise 7.4 of chapter 7-Integrals

### Exercise 7.4-Integrals

Integrate the functions in exercises 1 t0 23.

$\boldsymbol{Q1.\frac{3x^{2}}{x^{6}+1}}$

Ans. Integrating the functions

$\boldsymbol{\int \frac{3x^{2}}{x^{6}+1}dx}$

Let t = x³

$\frac{dt}{dx}=3x^{2}$

dt = 3x²dx

Substituting the value dt = 3x²dx and t = x³

$=\int \frac{dt}{t^{2}+1}$

$=tan^{-1}t+C$

Substituting back the value of t = x³

$=tan^{-1}x^{3}+C$

$\boldsymbol{Q2.\frac{1}{\sqrt{1+4x^{2}}}}$

Ans.We are given the function

${\frac{1}{\sqrt{1+4x^{2}}}}$

Integrating it

${\int \frac{dx}{\sqrt{1+4x^{2}}}}$

Let t = 2x

$\frac{dt}{dx}=2$

$dx=\frac{dt}{2}$

Substituting the value 2x=t and dx=dt/2

$=\frac{1}{2}\int \frac{dt}{\sqrt{1+t^{2}}}$

Using the formula for integration

$\int \frac{dx}{\sqrt{x^{2}+a^{2}}}=log\left | x+\sqrt{x^{2}+a^{2}} \right |$

Therefore

$=\int \frac{dt}{\sqrt{1+t^{2}}}=log\left | 1+\sqrt{1+t^{2}} \right |$

Substituting back the value of t=2x

$=log\left | 1+\sqrt{1+4x^{2}} \right |$

$\boldsymbol{Q3.\frac{1}{\sqrt{\left ( 2-x \right )^{2}+1}}}$

Ans. We are given

${\frac{1}{\sqrt{\left ( 2-x \right )^{2}+1}}}$

Let t= 2-x

$\frac{dt}{dx}=-1$

dx= -dt

Putting value of dx =-dt and 2-x= t in the integral

$-\int \frac{dt}{\sqrt{t^{2}+1}}$

Applying the integration formula

$\int \frac{dx}{\sqrt{x^{2}+a^{2}}}=log\left | x+\sqrt{x^{2}+a^{2}} \right |$

We have

$-\int \frac{dt}{\sqrt{t^{2}+1}}=-log\left | t+\sqrt{t^{2}+1} \right |$

Substituting back the value of t= 2-x, we get

$=-log\left | 2-x+\sqrt{\left ( 2-x \right )^{2}+1} \right |+C$

$=\frac{1}{log\left | 2-x+\sqrt{\left ( 2-x \right )^{2}+1} \right |}+C$

$\boldsymbol{Q4.\frac{1}{\sqrt{9-25x^{2}}}}$

Ans.

We are given

$\boldsymbol{\frac{1}{\sqrt{9-25x^{2}}}}$

Integrating it

$\int {\frac{dx}{\sqrt{9-25x^{2}}}}$

Let t= 5x

$\frac{dt}{dx}=5$

$dx=\frac{dt}{5}$

Substituting the value of dx =dt/5 and 5x = t

$=\frac{1}{5}\int \frac{dt}{\sqrt{3^{2}-t^{2}}}$

Applying the integration formula

$\int \frac{dx}{\sqrt{a^{2}-x^{2}}}=sin^{-1}\frac{x}{a}$

We have

$=\frac{1}{5}sin^{-1}\frac{t}{3}+C$

Substituting back the value of t=5x

$=\frac{1}{5}sin^{-1}\frac{5x}{3}+C$

### Class 12 Maths NCERT solutions of Exercise 7.4 of chapter 7-Integrals

$\boldsymbol{Q5.\frac{3x}{1+2x^{4}}}$

Ans. We are given

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

Integrating it

$\int {\frac{3x}{1+2x^{4}}}dx$

Let t= √2x²

$\frac{dt}{dx}=2\sqrt{2}x$

$dt=2\sqrt{2}xdx$

$xdx=\frac{dt}{2\sqrt{2}}$

Substituting  value of xdx ,we have

$=\frac{3}{2\sqrt{2}}\int \frac{dt}{1+t^{2}}$

Applying the integration formula

$\int \frac{dx}{a^{2}+x^{2}}=tan^{-1}\frac{x}{a}$

We get

$=\frac{3}{2\sqrt{2}}tan^{-1}t+C$

Substituting back the value of t=√2 x²

$=\frac{3}{2\sqrt{2}}tan^{-1}\left ( \sqrt{2} x^{2}\right )+C$

$\boldsymbol{Q6.\frac{x^{2}}{1-x^{3}}}$

We are given

$\boldsymbol{\frac{x^{2}}{1-x^{3}}}$

Integrating the given function

$\int \frac{x^{2}}{1-x^{6}}dx$

Let t = x³

$\frac{dt}{dx}=3x^{2}$

$x^{2}dx=\frac{dt}{3}$

Substituting the value x²dx = dt/3 and t= x³, we get

$= \frac{1}{3}\int \frac{dt}{1-t^{2}}$

Applying the integration formula

$= \int \frac{dx}{a^{2}-x^{2}}=\frac{1}{2}log\left | \frac{x+a}{x-a} \right |$

$=\frac{1}{3}.\frac{1}{2}log\left | \frac{1+t}{1-t} \right |+C$

$=\frac{1}{6}.log\left | \frac{1+t}{1-t} \right |+C$

Substituting back the value of t = x³

$=\frac{1}{6}log\left | \frac{1+x^{3}}{1-x^{3}} \right |+C$

.$\boldsymbol{Q7.\frac{x-1}{\sqrt{x^{2}-1}}}$

Ans.We are given the function

$\boldsymbol{\frac{x-1}{\sqrt{x^{2}-1}}}$

Integrating the function

$\int {\frac{x-1}{\sqrt{x^{2}-1}}}dx$

$=\int \frac{x}{\sqrt{x^{2}-1}}dx-\int \frac{dx}{\sqrt{x^{2}-1}}$

Solving integration of both fraction separately

Solution of first part

Let t = x² – 1

$\frac{dt}{dx}=2x\Rightarrow 2xdx=dt\Rightarrow xdx=\frac{dt}{2}$

Substituting the value of xdx = dt/2 and x² – 1 = t

$=\frac{1}{2}\int \frac{dt}{\sqrt{t}}$

$=\frac{1}{2}\int t^{-1/2}dt$

$=\frac{1}{2} .\frac{t^{1/2}}{1/2}=\frac{2}{2}\sqrt{t}=\sqrt{t}$

Substituting the value of  t = x² -1

We have

Integration of first part =$\sqrt{x^{2}-1}$

Putting this value as the integration of  first part and applying the following  integration formula for solving the second fraction

$\int \frac{dx}{\sqrt{x^{2}-a^{2}}}=log\left | x+\sqrt{x^{2}-a^{2}} \right |$

We have

$=\sqrt{x^{2}-1}-log\left | x+\sqrt{x^{2}-1} \right |+C$

$\boldsymbol{Q8.\frac{x^{2}}{\sqrt{x^{6}+a^{6}}}}$

Ans. We are given

$\boldsymbol{\frac{x^{2}}{\sqrt{x^{6}+a^{6}}}}$

Integrating the function

${\int \frac{x^{2}}{\sqrt{x^{6}+a^{6}}}}dx$

Let t= x³

$\frac{dt}{dx}=3x^{2}$

3x²dx =dt

$x^{2}dx=\frac{dt}{3}$

Substituting the value of x²dx = dt/3 and x³ = t

$\frac{1}{3}\int \frac{dt}{\sqrt{t^{2}+(a^{3})^{2}}}$

Applying the integration formula

$\because \int \frac{dx}{\sqrt{x^{2}+a^{2}}}=log\left | x+\sqrt{x^{2}+a^{2}} \right |$

$\frac{1}{3}\int \frac{dt}{\sqrt{t^{2}+(a^{3})^{2}}}=log\left | t +\sqrt{t^{2}+a^{6}}\right |+C$

Substituting back the value of t = x³

$=log\left | x^{3}+\sqrt{x^{6}+a^{6}} \right |+C$

### Class 12 Maths NCERT solutions of Exercise 7.4 of chapter 7-Integrals

$\boldsymbol{ Q9.{\frac{sec^{2}x}{\sqrt{tan^{2}x+4}}}}$

Ans. We are given

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

Integrating the given function

$\int {\frac{sec^{2}x}{\sqrt{tan^{2}x+4}}}dx$

Let t = tanx

$\frac{dt}{dx}=sec^{2}x$

dt = sec²x.dx

Substituting the value of sec²x.dx = dt and tanx=t

$\int \frac{dt}{\sqrt{t^{2}+2^{2}}}$

Applying the integration formula

$\int \frac{dx}{\sqrt{x^{2}+a^{2}}}=log\left | x+\sqrt{x^{2}+a^{2}} \right |$

$\int \frac{dx}{\sqrt{t^{2}+2^{2}}}=log\left | t+\sqrt{t^{2}+4} \right |+C$

Substituting back the value of t=tanx

$=log\left | tanx+\sqrt{tan^{2}x+4} \right |+C$

$\boldsymbol{Q10.\frac{dx}{\sqrt{x^{2}+2x+2}}}$

Ans. We are given

$\frac{dx}{\sqrt{x^{2}+2x+2}}$

Integrating the given function

$\int \frac{dx}{\sqrt{x^{2}+2x+2}}$

Let t = x+1 ,we can rewrite the denominator x² +2x+2= (x+1)² +1

$\int {\frac{dx}{\sqrt{\left ( x+1 \right )^{2}+1}}}$

$\frac{dt}{dx}=1$

dt = dx

$\int {\frac{dt}{\sqrt{t^{2}+1}}}$

Applying the integration formula

$\int {\frac{dx}{\sqrt{x^{2}+a^{2}}}}=log\left | x+\sqrt{x^{2}+a^{2}} \right |$

$\therefore \int {\frac{dt}{\sqrt{t^{2}+1}}}=log\left | t+\sqrt{t^{2}+1} \right |+C$

Substituting back the value of t = x+1

$=log\left | x+1+\sqrt{(x+1)^{2}+1} \right |+C$

$=log\left | x+1+\sqrt{x^{2}+2x+2} \right |+C$

$\boldsymbol{Q11.\frac{1}{9x^{2}+6x+5}}$

Ans.We are given the function

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

Integrating it

$\int {\frac{dx}{9x^{2}+6x+5}}$

Rewriting the denominator of the given function

9x² +6x + 1 +4 = (3x+1)² +2²

$\int {\frac{dx}{\left ( 3x+1 \right )^{2}+2^{2}}}$

Let t = 3x +1

$\frac{dt}{dx}=3$

$dx=\frac{dt}{3}$

Substituting the value 3x+1=t and dx= dt/3

$\frac{1}{3}\int \frac{dt}{t^{2}+2^{2}}$

Applying the integration formula

$\int \frac{dx}{x^{2}+a^{2}}=\frac{1}{a}tan^{-1}\frac{x}{a}$

We have

$\frac{1}{3}\int \frac{dt}{t^{2}+2^{2}}=\frac{1}{3}.\frac{1}{2}tan^{-1}\left ( \frac{t}{2} \right )$

$=\frac{1}{6}tan^{-1}\left ( \frac{t}{2} \right )+C$

Substituting back the value of t=3x+1

$=\frac{1}{6}tan^{-1}\frac{3x+1}{2}+C$

$\boldsymbol{Q12.\frac{1}{\sqrt{7-6x-x^{2}}}}$

Ans. We are given

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

Converting the terms of denominator into a complete square form

7 -6x -x² = -(x² +6x -7+3² -3²)= -(x+3)² +16= 4² -(x+3)²

So,we have

$=\frac{1}{\sqrt{4^{2}-\left ( x+3 \right )^{2}}}$

Now,integrating the function

$=\int \frac{dx}{\sqrt{4^{2}-\left ( x+3 \right )^{2}}}$

Let t= x+3

$\frac{dt}{dx}=1$

dt = dx

Substituting the value of dx=dt and x+3= t

$\int \frac{dt}{\sqrt{4^{2}-t^{2}}}$

Applying the formula of integration

$\int \frac{dx}{\sqrt{a^{2}-x^{2}}}=sin^{-1}\frac{x}{a}$

$\therefore \int \frac{dt}{\sqrt{4^{2}-t^{2}}}=sin^{-1}\frac{t}{4}+C$

Substituting back the value of t = x +3

$=sin^{-1}\frac{x+3}{4}+C$

### Class 12 Maths NCERT solutions of Exercise 7.4 of chapter 7-Integrals

$\boldsymbol{Q13.\frac{1}{\sqrt{\left ( x-1 \right )\left ( x-2 \right )}}}$

Ans. We are given the function

${\frac{1}{\sqrt{\left ( x-1 \right )\left ( x-2 \right )}}}$

Integrating the given function

Converting the terms of denominator into complete square form

(x-1)(x-2)=x²-3x +2 +(3/2)² -(3/2)²

= (x-3/2)²-(1/2)²

So,we have

$\frac{1}{\left ( x-3/2 \right )^{2}-\left ( 1/2 \right )^{2}}$

Now, integrating it

$\int \frac{dx}{\sqrt{\left ( x-3/2 \right )^{2}-\left ( 1/2 \right )^{2}}}$

$Let\: t=x-\frac{3}{2}$

$\frac{dt}{dx}=1$

dt =dx

Substituting x-3/2= t and dx= dt in the integral

$=\int \frac{dt}{\sqrt{t^{2}-\left ( 1/2 \right )^{2}}}$

Using the integration formula

$\int \frac{dx}{\sqrt{x^{2}-a^{2}}}=log\left | x+\sqrt{x^{2}-a^{2}} \right |$

So, we have

$=\int \frac{dt}{\sqrt{t^{2}-\left ( 1/2 \right )^{2}}}=log\left | t+\sqrt{t^{2}-\left ( 1/2 \right )^{2}} \right |+C$

Substituting back the value of t = x -3/2, we have

$=log\left | x-3/2+\sqrt{\left ( x-3/2 \right )^{2}-\left ( 1/2 \right )^{2}} \right |+C$

$=log\left | x-3/2+\sqrt{x^{2}-3x+2} \right |+C$

$\boldsymbol{Q14.\frac{1}{\sqrt{8+3x-x^{2}}}}$

Converting the denominator 8 + 3x -x² into the terms of complete square

8 + 3x -x²

$\Rightarrow -\left [ x^{2} -3x-8+\left ( \frac{3}{2} \right )^{2}-\left ( \frac{3}{2} \right )^{2}\right ]$

$\Rightarrow -\left [ \left ( x-\frac{3}{2} \right )^{2}-\frac{41}{4} \right ]=\left [ \frac{41}{4} -\left ( x-\frac{3}{2} \right )^{2}\right ]$

Now, we have

$\frac{1}{\sqrt{\left ( \sqrt{41}/2 \right )^{2}-\left ( x-3/2 \right )^{2}}}$

Now,integrating the function

$\int \frac{dx}{\sqrt{\left ( \sqrt{41}/2 \right )^{2}-\left ( x-3/2 \right )^{2}}}$

Let t= x -3/2

$\frac{dt}{dx}=1$

dt =dx

Substituting the value dx=dt and t= x -3/2

$\int \frac{dt}{\sqrt{\left ( \sqrt{41}/2 \right )^{2}-t^{2}}}$

Applying the integration formula

$\int \frac{dx}{\sqrt{a^{2}-x^{2}}}=sin^{-1}\frac{x}{a}$

So, we have

$\int \frac{dt}{\sqrt{\left ( \sqrt{41}/2 \right )^{2}-t^{2}}}=sin^{-1}\frac{t}{\sqrt{41}/2}+C$

Substituting back the value of t = x-3/2

$=sin^{-1}\frac{2x-3}{\sqrt{41}}+C$

$\boldsymbol{Q15.\frac{1}{\sqrt{\left ( x-a \right )\left ( x-b \right )}}}$

Ans. We are given

${\frac{1}{\sqrt{\left ( x-a \right )\left ( x-b \right )}}}$

Converting the terms of denominator into the form of complete square

(x -a)(x-b)  = x² – (a +b)x + ab=x² – (a +b)x + ab +[(a+b)/2]² -[(a+b)/2]²

$\Rightarrow \left ( x-\frac{a+b}{2} \right )^{2}-\left ( \frac{a-b}{2} \right )^{2}$

${\frac{1}{\sqrt{\left ( x-a \right )\left ( x-b \right )}}}=\frac{1}{\sqrt{\left ( x-\frac{a+b}{2} \right )^{2}-\left ( \frac{a-b}{2} \right )^{2}}}$

Integrating it

$\int \frac{dx}{\sqrt{\left ( x-\frac{a+b}{2} \right )^{2}-\left ( \frac{a-b}{2} \right )^{2}}}$

$Let\: t= x-\frac{a+b}{2}$

$\frac{dt}{dx}=1$

dx =dt

$\int \frac{dt}{\sqrt{t^{2}-\left ( \frac{a-b}{2} \right )^{2}}}$

$\int \frac{dt}{\sqrt{t^{2}-\left ( \frac{a-b}{2} \right )^{2}}}=log\left | t+\sqrt{t^{2}-(\frac{a-b}{2})^{2}} \right |+C$

Substituting the value of    $t=x-\frac{a+b}{2}$

$=log\left | x-\frac{a+b}{2}+\sqrt{\left ( x-\frac{a+b}{2} \right )^{2}-\left ( \frac{a-b}{2} \right )^{2}} \right |+C$

$=log\left | x-\frac{a+b}{2}+\sqrt{\left ( x-a \right )\left ( x-b \right )} \right |+C$

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