Question

$\int \frac{1}{x\sqrt{1+x^n}}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{dx}{x\sqrt{1+x^n}} \\ =\int \frac{x^{n-1}dx}{x^{n-1}{x}^1\sqrt{1+x^n}} \\ =\int \frac{x^{n-1}dx}{x^n\sqrt{1+x^n}} \\ \mathrm{Putting}{x}^n=t \\ \Rightarrow n{x}^{n-1}dx=dt \\ \Rightarrow x^{n-1}dx=\frac{dt}{n} \\ \therefore I=\frac{1}{n}\int \frac{dt}{t\sqrt{1+t}} \\ \mathrm{let}1+t=p^2 \\ \Rightarrow dt=2pdp \\ \therefore I=\frac{1}{n}\int \frac{2pdp}{\left(p^2-1\right)p} \\ =\frac{2}{n}\int \frac{dp}{p^2-1^2} \\ =\frac{2}{n}\times \frac{1}{2}\log \left|\frac{p-1}{p+1}\right|+C \\ =\frac{1}{n}\log \left|\frac{\sqrt{1+t}-1}{\sqrt{1+t}+1}\right|+C \\ =\frac{1}{n}\log \left|\frac{\sqrt{1+x^n}-1}{\sqrt{1+x^n}+1}\right|+C \end{gathered}$$