Question
Evaluate: $\int\frac{2}{\text{(1-x)(1+x)}^{2}}\text{dx}$

Answer

$\frac{2}{\text{(1-x)}\text{(1+x}^{2})}=\frac{\text{A}}{\text{1-x}}+\frac{\text{Bx+C}}{\text{1+x}^{2}}$
2 = A(1 + x2) + (Bx + C) (1 – x)
$\Rightarrow$ 0 = A – B, B – C = 0, A + C = 2 $\Rightarrow$ A = B = C = 1
$\therefore\int\frac{2}{\text{(1-x)(1+x}^{2})}\text{dx}=\int\frac{1}{\text{1-x}}\text{dx}+\frac{\text{x+1}}{\text{x}^{2}+{1}}\text{dx}$
= – log |1 – x| + $\frac{1}{2}$ log (x2 + 1) + tan-1 x + c.

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