Question

Integrate the following with respect to x.

`(4x^2 + 2x + 6)/((x + 1)^2(x - 3))`

Answer 1

By partial fraction

`(4x^2 + 2x + 6)/((x +1)^2(x - 3)) = "A"/((x - 3)) + "B"/((x + 1)) + "C"/(x + 1)^2`

`(4x^2 + 2x + 6)/((x +1)^2(x - 3)) = ("A"(x + 1)^2 + "B"(x - 3)(x + 1) + "c"(x - 3))/((x - 3)(x + 1)^2`

⇒ 4x² + 2x + 6 = A(x + 1)² + B(x – 3)(x + 1) + C(x – 3)

Put x = 3

4(9) + 2(3) + 6 = A(4)²

36 + 6 + 6 = 16A 

⇒ A = `48/16`

A = 3

Put x = – 1

4(1) + 2(– 1) + 6 = C(– 4)

4 – 2 + 6 = – 4C

⇒ – 4c = 8

C = – 2

Put x = 0

6 = A(1) + B(– 3) + C(– 3)

6 = 3(1) – 3B + (– 2)(– 3)

6 = 3 – 3B + 6

⇒ 3B = 3

B = 1

`(4x^2 + 2x + 6)/((x + 1)^2(x - 3)) = 3/((x - 3)) + 1/((x + 1)) - 2/(x + 1)^2`

`int (4x^2 + 2x + 6)/((x + 1)^2(x - 3))  "d"x = int [3/((x - 3)) + 1/((x + 1)) - 2(x + 1)^-2]  "d"x`

= `3 log |x - 3| + log|x + 1| - 2 (x + 1)^(-2 + 1)/((-2 + 1)) + "C"`

= `3 log|x - 3| + log |x + 1| + 2/((x + 1)) + "C"`