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Class 12 Maths NCERT Solutions for Chapter 7 Integrals Exercise 7.4

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. 
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