Evaluate: ∫ 2/(1 - x)(1 + x²) dx

Let $\frac{2}{(1-x)(1+x^2)} = \frac{A}{1-x} + \frac{Bx+C}{1+x^2}$. $2 = A(1+x^2) + (Bx+C)(1-x)$ Put $x = 1$: $A = 1$. Comparing $x^2$: $0 = A - B$, so $B = 1$. Comparing constants: $2 = A + C$, so $C = 1$. $\int \frac{2}{(1-x)(1+x^2)}dx = \int\frac{dx}{1-x} + \int\frac{x+1}{1+x^2}dx$ $= -\log|1-x| + \frac{1}{2}\log(1+x^2) + \tan^{-1}x + C$ Hence the result is $-\log|1-x| + \frac{1}{2}\log(1+x^2) + \tan^{-1}x + C$.