**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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**Class 12 Maths NCERT solutions of Chapter 7 Integrals**

**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.

Ans. Integrating the functions

Let t = x³

dt = 3x²dx

Substituting the value dt = 3x²dx and t = x³

Substituting back the value of t = x³

Ans.We are given the function

Integrating it

Let t = 2x

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

Using the formula for integration

Therefore

Substituting back the value of t=2x

Ans. We are given

Let t= 2-x

dx= -dt

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

Applying the integration formula

We have

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

Ans.

We are given

Integrating it

Let t= 5x

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

Applying the integration formula

We have

Substituting back the value of t=5x

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

Ans. We are given

Integrating it

Let t= √2x²

Substituting value of xdx ,we have

Applying the integration formula

We get

Substituting back the value of t=√2 x²

We are given

Integrating the given function

Let t = x³

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

Applying the integration formula

Substituting back the value of t = x³

.

Ans.We are given the function

Integrating the function

Solving integration of both fraction separately

Solution of first part

Let t = x² – 1

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

Substituting the value of t = x² -1

We have

Integration of first part =

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

We have

Ans. We are given

Integrating the function

Let t= x³

3x²dx =dt

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

Applying the integration formula

Substituting back the value of t = x³

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

Ans. We are given

Integrating the given function

Let t = tanx

dt = sec²x.dx

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

Applying the integration formula

Substituting back the value of t=tanx

Ans. We are given

Integrating the given function

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

dt = dx

Applying the integration formula

Substituting back the value of t = x+1

Ans.We are given the function

Integrating it

Rewriting the denominator of the given function

9x² +6x + 1 +4 = (3x+1)² +2²

Let t = 3x +1

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

Applying the integration formula

We have

Substituting back the value of t=3x+1

Ans. We are given

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

Now,integrating the function

Let t= x+3

dt = dx

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

Applying the formula of integration

Substituting back the value of t = x +3

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

Ans. We are given the function

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

Now, integrating it

dt =dx

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

Using the integration formula

So, we have

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

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

8 + 3x -x²

Now, we have

Now,integrating the function

Let t= x -3/2

dt =dx

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

Applying the integration formula

So, we have

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

Ans. We are given

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

Integrating it

dx =dt

Substituting the value of

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