Let $I = \frac{1}{x-x^3}dx$ $\frac{1}{x-x^3} = \frac{1}{x(1+x)(1-x)}$ $= \frac{A}{x} + \frac{B}{1+x} + \frac{C}{1-x}$ ⇒ 1 = A(1 + x)(1 – x) + Bx (1 – x) + Cx(1 + x) ...(i)Putting x = 0 in (i), we get, 1 = A(1 + 0) (1 – 0) ⇒ A = 1Putting x = –1 in (i), we get 1 = B(–1)(1 + 1) ⇒ B $= -\frac{1}{2}$Putting x = 1 in (i), we get 1 = C(1)(1 + 1) ⇒ C $= \frac{1}{2}$$\frac{1}{x-x^3} = \frac{1}{x} - \frac{1}{2(1+x)} + \frac{1}{2(1-x)}$$\therefore \int \frac{1}{x-x^3}\,dx$ $= \int \frac{1}{x}\,dx - \frac{1}{2} \int \frac{1}{1+x}\,dx + \frac{1}{2} \int \frac{1}{1-x}\,dx$$= \log |x| - \frac{1}{2} \log |1+x| - \frac{1}{2} \log |1-x| + C$$= \log |x| - \frac{1}{2} \log |1+x^2| + C$$= \frac{1}{2} \log \left| \frac{x^2}{1-x^2} \right| + C$