Question 1

Question 1:

Accepted Answer

Let I = Let 3x + 4 = A[(x2 + 6x + 5)] + B = A(2x + B) + B ∴ 3x + 4 = 2Ax + (6A + B) Comparing the coefficient of x and constant on both sides, we get 2A = 3 and 6A + B = 4 ∴ A = and 6() + B = 4 ∴ B = -5 3x + 4 = (2x + 6) – 5

Question 2

Question 2:

Accepted Answer

Let I = Let 2x + 1 = A[(x2 + 4x – 5)] + B 2x + 1 = A(2x + 1) + B ∴ 2x + 1 = 2Ax + (4A + B) Comparing the coefficient of x and constant on both sides, we get 4A = 2 and 4A + B = 4 ∴ A = and 6() + B = 4 ∴ B = -5 ∴ 2x + 1 = (2x + 1) – 5

Question 3

Question 3:

Accepted Answer

Let I = Let 2x+ 3 = A[(2x2 + 3x – 1)] + B 2x + 1 = A(4x + 3) + B ∴ 2x + 1 = 4Ax + (3A + B) Comparing the coefficient of x and constant on both sides, we get 4A = 2 and 3A + B = 3 ∴ A = and 3() + B = 3 ∴ B = ∴ 2x + 3 = (4x + 3) +

Question 4

Question 4:

Accepted Answer

Let I = Let 3x + 4 = A[(2x2 + 2x + 1)] + B ∴ 3x + 4 = A (4x + 2) + B ∴ 3x + 4 = 4Ax + (2A + B) Comparing the coefficient of x and the constant on both the sides, we get 4A = 3 and 2A + B = 4 ∴ A = and 2() + B = 4 ∴ B = ∴ 3x + 4 = (4x + 2) +

Question 5

Question 5:

Accepted Answer

Let I = Let 7x + 3 = A[(3 + 2x – x2)] + B 7x + 3 = A(2 – 2x) + B ∴ 7x + 3 = -2Ax + (2A + B) Comparing the coefficient of x and constant on both the sides, we get -2A = 7 and 2A + B = 3

Question 6

Question 6:

Accepted Answer

Comparing the coefficients of x and constant term on both sides, we get

Question 7

Question 7:

Accepted Answer

Step-by-step solution provided.

Question 8

Question 8:

Accepted Answer

Step-by-step solution provided.

Question 9

Question 9:

Accepted Answer

Step-by-step solution provided.

Chapter 3 Indefinite Integration Ex 3.2(C) Solutions