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