Question

Evaluate: `int (3x^2 + 4x - 5)/((x^2 - 1)(x + 2)) dx`

Answer 1

`int (3x^2 + 4x - 5)/((x^2 - 1)(x + 2)) dx`

`int (3x^2 + 4x - 5)/((x + 1)(x - 1)(x + 2)) dx`   ...[a² – b² = (a + b)(a – b)]

Let `(3x^2 + 4x - 5)/((x + 1)(x - 1)(x + 2)) = A/(x + 1) + B/(x - 1) + C/(x + 2)`

`(3x^2 + 4x - 5)/(\cancel((x + 1))\cancel((x - 1))(x + 2)) = (A(x - 1)(x + 2) + B(x + 1)(x + 2) + C(x + 1)(x - 1))/(\cancel((x + 1))\cancel((x - 1))(x + 2))`

3x² + 4x – 5 = A(x² + x – 2) + B(x² + 3x + 2) + C(x² – 1)

3x² + 4x – 5 = x²(A + B + C) + x(A + 3B) – 2A + 2B – C

A + B + C = 3   ...(1)

A + 3B = 4   ...(2)

–2A + 2B – C = –5

`A - B + C/2 = 5/2`   ...(3)

Subtracting (2) from (1)

–2B + C = –1

2B – C = 1   ...(4)

Subtracting (3) from (2)

`4B - C/2 = 3/2`   ...(5)

Multiplying (5) by 2

8B – C = 3   ...(6)

Subtracting (4) from (6)

6B = 2

B = `2/6`

B = `1/3`

`8(1/3) - C = 3`

`C = 8/3 - 3`

`C = (8 - 9)/3`

C = `(-1)/3`

`A + \cancel((1/3)) + \cancel(((-1)/3)) = 3`

A = 3

`int 3/(x + 1) * dx + int (1/3)/(x - 1) * dx + int ((-1)/3)/(x + 2) * dx`

`3 int 1/(x + 1) * dx + 1/3 int 1/(x - 1) * dx  - 1/3 int 1/(x + 2) * dx`

`3 * ln (x + 1) + 1/3 ln (x - 1) - 1/3 ln (x + 2) + C`