Question

Find: `int (2x + 1)/((x + 1)^2(x - 1)) dx`

Answer 1

Let I = `int (2x + 1)/((x + 1)^2(x - 1)) dx`

Let `(2x + 1)/((x +1)^2 (x - 1)) = A/((x + 1)) + b/((x + 1)^2) + C/((x - 1))` 

2x + 1 = A(x + 1)(x − 1) + B(x − 1) + C(x + 1)²

2x + 1 = A(x² − 1) + Bx − B + C(x² + 2x +1)

On comparing, we get

A + C = 0; B + 2C = 2 and − A − B + C = 1

On solving, we get

A = `(-3)/4`, B = `1/2` and C = `3/4`

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

Thus, `int (2x + 1)/((x + 1)^2(x - 1)) dx`

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

= `-3/4 log |x + 1| + 1/2 I_1 + 3/4 log |x -1| + C_1`

How, `I_1 = int 1/(x + 1)^2 dx`

let x + 1 = u ⇒ dx + du

∴ `I_1 = int 1/u^2 du = u^(-2+1)/(-2 + 1) + C_2`

= `-1/u + C_2`

= `- 1/(x + 1) + C_2`

Therefore, `int (2x + 1)/((x + 1)^2(x - 1))dx`

= `-3/4 log |x + 1| - 1/(2(x + 1)) + 3/4 log |x - 1| + C`