# NCERT solution class 12 chapter 1 Integrals exercise 1.4 mathematics part 2

## EXERCISE 1.4

#### Question 1: Let x3 = t

∴ 3x2 dx = dt #### Question 2: Let 2x = t

∴ 2dx = dt #### Question 3: Let 2 − t

⇒ −dx = dt #### Question 4: Let 5x = t

∴ 5dx = dt #### Question 5:  #### Question 6: Let x3 = t

∴ 3x2 dx = dt #### Question 7:  From (1), we obtain #### Question 8: Let x3 = t

⇒ 3x2 dx = dt #### Question 9: Let tan x = t

∴ sec2x dx = dt #### Question 10:  #### Question 11:

19×2+6x+5

∫19×2+6x+5dx=∫13x+12+22dx

Let (3x+1)=t

3 dx=dt

⇒∫13x+12+22dx=13∫1t2+22dt

=13×2tan-1t2+C

=16tan-13x+12+C

#### Question 12:  #### Question 13:  #### Question 14:  #### Question 15:  #### Question 16:  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 #### Question 17:  Equating the coefficients of x and constant term on both sides, we obtain From (1), we obtain From equation (2), we obtain #### Question 18:  Equating the coefficient of x and constant term on both sides, we obtain    Substituting equations (2) and (3) in equation (1), we obtain #### Question 19:   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  Substituting equations (2) and (3) in (1), we obtain #### Question 20:  Equating the coefficients of x and constant term on both sides, we obtain  Using equations (2) and (3) in (1), we obtain #### Question 21:  Let x2 + 2x +3 = t

⇒ (2x + 2) dx =dt Using equations (2) and (3) in (1), we obtain  #### Question 22:  Equating the coefficients of x and constant term on both sides, we obtain   Substituting (2) and (3) in (1), we obtain #### Question 23:  Equating the coefficients of x and constant term, we obtain  Using equations (2) and (3) in (1), we obtain #### Question 24: 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 Hence, the correct answer is B.

#### Question 25: equals

A. B. C. D.   