Let $\frac{1}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}$ $\Rightarrow 1 = A(x^2 + 1) + (Bx + C)(x)$...(i)Let $x = 0, 1 = A \Rightarrow A = 1$Comparing coefficients of $x^2$ in (i) on both sides, we get $0 = A + B \Rightarrow B = -1$Comparing coefficients of x in (i) on both sides, we get, $C = 0$$\therefore \int \frac{dx}{x(x^2+1)} = \int \left[ \frac{1}{x} + \frac{-x}{x^2+1} \right] dx$$= \log{|x|} - \frac{1}{2} \log(x^2+1) + C$