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

## EXERCISE 1.4

Let x3 = t

∴ 3x2 dx = dt

Let 2x = t

∴ 2dx = dt

Let 2 − t

⇒ −dx = dt

Let 5x = t

∴ 5dx = dt

Let x3 = t

∴ 3x2 dx = dt

#### Question 7:

From (1), we obtain

Let x3 = t

⇒ 3x2 dx = dt

Let tan x = t

∴ sec2x dx = dt

#### 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 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.

equals

A.

B.

C.

D.