# Class 12 Maths NCERT Solutions for Chapter 7 Integrals Exercise 7.4

### Integrals Exercise 7.4 Solutions

1. Integrate the functions 3x2 /(x6 + 1)

Solution

Let x3 = t
⇒ 3x2 dx = dt

= tan-1 t + C
= tan-1 (x3 ) + C

2. Integrate the functions 1/√(1 + 4x2 )

Solution

Let 2x = t
⇒ 2dx = dt

3. Integrate the functions 1/√[(2 - x)2 + 1]

Solution

Let 2 - x = t
⇒ -dx = dt

4. Integrate the functions 1/√(9 - 25x2 )

Solution

Let 5x = t
⇒ 5dx = dt

5. Integrate the functions 3x/(1 + 2x4 )

Solution

Let √2x2 = t
∴ 2√2x dx = dt

6. Integrate the functions x2 /(1 - x6 )

Solution

Let x3 = t
⇒ 3x2 dx = dt

7. Integrate the functions (x - 1)/√(x2 - 1)

Solution

8. Integrate the functions x2 /√(x6 + a6 )

Solution

Let x3 = t
⇒ 3x2 dx = dt

9. Integrate the functions sec2 x/√(tan2 x + 4)

Solution

Let tan x = t
⇒ sec2 x dx = dt

10. Integrate the functions 1/√(x2 + 2x +2)

Solution

11. Integrate the function 1/√(9x2 + 6x + 5)

Solution

12. Integrate the functions 1/√(7 - 6x - x2 )

Solution

7 - 6x - x2 can be written as  7 - (x2 + 6x + 9 - 9).
Therefore,
7 - (x2 + 6x + 9 - 9)
= 16 - (x2 + 6x + 9)
= 16 - (x + 3)2
= (4)2 - (x + 3)2

13. Integrate the functions 1/√(x - 1)(x - 2)

Solution

(x - 1)(x - 2) can be written as x2 - 3x + 2.
Therefore,
x2 - 3x + 2
= x2 - 3x + 9/4 - 9/4 + 2

14. Integrate the functions  1/√(8 + 3x - x2 )

Solution

8 + 3x - x2 can be written as 8 - (x2 - 3x + 9/4 - 9/4).
Therefore,

15. Integrate the functions 1/√[(x- a)(x - b)]

Solution

(x - a)(x - b) can be written as x2 - (a + b)x + ab.
Therefore,
x2 - (a + b)x + ab

16. Integrate the functions (4x + 1)/√(2x2 + x - 3)

Solution

⇒ 4x + 1 = A(4x + 1) + B
⇒ 4x + 1 = 4Ax + A + B
Equating the coefficients of x and constant term on both sides, we obtain
4A = 4 ⇒ A = 1
A + B = 1 ⇒ B = 0
Let 2x2 + x − 3 = t
∴ (4x + 1) dx = dt

17. Integrate the functions (x + 2)/√(x2 - 1)

Solution

⇒ x + 2 = A(2x) + B
Equating the coefficients of x and constant term on both sides, we obtain
2A = 1 ⇒ A = 1/2
B = 2
From (1), we obtain

18. Integrate the functions (5x - 2)/(1 + 2x + 3x2 )

Solution

⇒ 5x - 2 = A(2 + 6x) + B
Equating the coefficient of x and constant term on both sides, we obtain
5 = 6A ⇒ A = 5/6
2A + B = -2 ⇒ B = -11/3

Let 1 + 2x + 3x2 = t
⇒ (2 + 6x) dx = dt
∴ I1 = ∫dt/t
I1 = log|t|
I1 = log|1 + 2x + 3x2 |     .....(2)

19. Integrate the functions (6x + 7)/√[(x - 5)(x - 4)] .

Solution

⇒ 6x + 7 = A(2x - 9) + B
Equating the coefficients of x and constant term, we obtain
2A = 6 ⇒ A = 3
−9A + B = 7 ⇒ B = 34
∴ 6x + 7 = 3 (2x − 9) + 34
x2 - 9x + 20 can be written as x2 - 9x + 20 + 81/4 - 81/4 .
Therefore,
x2 - 9x + 20 + 81/4 - 81/4

20.  Integrate the functions (x + 2)/√(4x - x2 )

Solution

⇒ x + 2 = A(4 - 2x) + B
Equating the coefficients of x and constant term of both sides, we obtair
-2A = 1 ⇒ A = -1/2
4A + B = 2 ⇒ B = 4
⇒ (x + 2) = -1/2(4 - 2x) + 4

21. Integrate the functions (x + 2)/√(x2 + 2x + 3)

Solution

Let x2 + 2x + 3 = t
⇒ (2x + 2)dx = dt

22. Integrate the functions (x + 3)/(x2 - 2x - 5)

Solution

(x + 3) = A(2x - 2) + B
Equating the coefficients of x and constant term on both sides, we obtain
2A = 1 ⇒ A = 1/2
-2A + B = 3 ⇒ B = 4
∴(x + 3) = (1/2)(2x - 2) + 4
Let x2 - 2x - 5 = t
⇒ (2x -2)dx = dt
⇒ I1 = ∫(dt/t) = log|t| = log|x2 - 2x - 5|   .....(2)

23. Integrate the functions (5x + 3)/√(x2 + 4x + 10)

Solution

⇒ 5x + 3 = A(2x + 4) + B
Equating the coefficients of x and constant term, we obtain
2A = 5 ⇒ A = 5/2
4A + B = 3 ⇒ B = -7

Using equations (2) and (3) in (1) we obtain

24. Choose the correct answer int. dx/(x2 + 2x + 2) equals
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

Solution

Hence, the correct answer is B.

25. Choose the correct answer ∫dx/√(9x - 4x2 ) equals
(A) (1/9)sin-1 [(9x - 8)/8] + C
(B) (1/2)sin-1 [(8x - 9)/9] + C
(C) (1/3)sin-1 [(9x - 8)/8] + C
(D) (1/2)sin-1 [(9x - 8)/9] + C

Solution

Hence, the correct answer is B.