Question:
Evaluate $\int \frac{1}{x \sqrt{1+x^{n}}} d x$
Solution:
Let, $\sqrt{1+x^{n}}=t$
Differentiate both side with respect to $\mathrm{t}$
$\frac{n x^{n-1}}{2 \sqrt{1+x^{n}}} \frac{d x}{d t}=1 \Rightarrow \frac{d x}{x \sqrt{1+x^{n}}}=\frac{2 d t}{n\left(t^{2}-1\right)}$
$y=\int \frac{2}{n\left(t^{2}-1\right)} d t$
Use formula $\int \frac{1}{t^{2}-a^{2}} d t=\frac{1}{2 a} \ln \left(\frac{t-a}{t+a}\right)$
$y=\frac{1}{n} \ln \left(\frac{t-1}{t+1}\right)+c$
Again put $t=\sqrt{1+x^{n}}$
$y=\frac{1}{n} \ln \left(\frac{\sqrt{1+x^{n}}-1}{\sqrt{1+x^{n}}+1}\right)+c$