Question

Evaluate: 

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

Answer 1

∴ `2/((1 - x)(1 + x^2)) = "A"/(1 - x) + ("B"x + "C")/(1 + x^2)`  ...(1)

`\implies` 2 = A(1 + x²) + (Bx + C)(1 – x)

`\implies` 2 = A(1 + x²) + (– Bx² + Bx – Cx + C)

`\implies` 2 = (A – B)x² + (B – C)x + A + C

Equating the coefficients of like terms.

A – B = 0,

B – C = 0

and A + C = 2

On adding,

2A = 2

`\implies` A = 1

∴ B = A = 1

and C = B = 1

By (1) `2/((1 - x)(1 + x^2)) = 1/(1 - x) + (x + 1)/(1 + x^2)`

Now integrating

`int 2/((1 - x)(1 + x^2))"d"x = int 1/(1 - x)"d"x + int (x + 1)/(1 + x^2)"d"x`

= `int 1/(1 - x)"d"x + int x/(1 + x^2)"d"x + int 1/(1 + x^2)"d"x`

= `-log(1 - x) + 1/2log(1 + x^2) + tan^-1x + "C"`.