Let $I=\int \frac{2}{(1-x)(1+x^2)}dx$ Let $\frac{2}{(1-x)(1+x^2)} = \frac{A}{1-x}+\frac{Bx+C}{1+x^2}$ $\Rightarrow 2 = A(1 + x^2) + (Bx + C) (1 - x)$ ...(i)Putting x = 1 in (i), we get 2 = A(1 + 1) ⇒ A = 1Comparing coefficients of $x^2$ on both sides of (i), we get 0 = A – B ...(ii)Comparing coefficients of constant terms on both sides of (i), we get 2 = A + C ...(iii)Put A = 1 in (ii), we get B = 1Put A = 1 in (iii), we get C = 1$\therefore \frac{2}{(1-x)(1+x^2)} = \frac{1}{1-x} + \frac{x+1}{1+x^2}$