Let $I=\int \frac{1}{x(x^4-1)}\,dx$ $=\frac{1}{4}\int \frac{4x^3\,dx}{x^4(x^4-1)}$ Put $x^4 = t \Rightarrow 4x^3\,dx = dt$ $\therefore I=\frac{1}{4}\int \frac{dt}{t(t-1)}$ Let $\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}$ $\Rightarrow 1 = A(t-1) + Bt$...(i)Putting t = 0 in (i), we get 1 = A (–1) ⇒ A = –1Putting t = 1 in (i), we get 1 = B(1) ⇒ B = 1 $\therefore \frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{t-1}$$\therefore I=\frac{1}{4}\int \left( \frac{-1}{t}+\frac{1}{t-1} \right) dt$$=\frac{1}{4}\left[ -\log |t| + \log |t-1| \right] + C$$=\frac{1}{4}\log \left| \frac{t-1}{t} \right| + C$$=\frac{1}{4}\log \left| \frac{x^4-1}{x^4} \right| + C$