Show that

$\frac{1}{x\left(x^{n}+1\right)}$

Multiplying numerator and denominator by $x^{n-1}$, we obtain

$\frac{1}{x\left(x^{n}+1\right)}=\frac{x^{n-1}}{x^{n-1} x\left(x^{n}+1\right)}=\frac{x^{n-1}}{x^{n}\left(x^{n}+1\right)}$

Let $x^{n}=t \Rightarrow x^{n-1} d x=d t$

$\therefore \int \frac{1}{x\left(x^{n}+1\right)} d x=\int \frac{x^{n-1}}{x^{n}\left(x^{n}+1\right)} d x=\frac{1}{n} \int \frac{1}{t(t+1)} d t$

Let $\frac{1}{t(t+1)}=\frac{A}{t}+\frac{B}{(t+1)}$

$1=A(1+t)+B t$    ...(1)

Substituting t = 0, −1 in equation (1), we obtain

$A=1$ and $B=-1$

$\therefore \frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{(1+t)}$

$\Rightarrow \int \frac{1}{x\left(x^{n}+1\right)} d x=\frac{1}{n} \int\left\{\frac{1}{t}-\frac{1}{(t+1)}\right\} d x$

$=\frac{1}{n}[\log |t|-\log |t+1|]+\mathrm{C}$

$=-\frac{1}{n}\left[\log \left|x^{n}\right|-\log \left|x^{n}+1\right|\right]+\mathrm{C}$

$=\frac{1}{n} \log \left|\frac{x^{n}}{x^{n}+1}\right|+\mathrm{C}$