Question

$\int x\sqrt{x^4+1}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int x\sqrt{x^4+1}dx \\ =\int x\sqrt{{\left(x^2\right)}^2+1}dx \\ \mathrm{Putting}{x}^2=t \\ \Rightarrow 2xdx=dt \\ \Rightarrow xdx=\frac{dt}{2} \\ \therefore I=\frac{1}{2}\int \sqrt{t^2+1}dt \\ =\frac{1}{2}\int \sqrt{t^2+1^2}dt \\ =\frac{1}{2}\left[\frac{t}{2}\sqrt{t^2+1}+\frac{1^2}{2}\log \left|t+\sqrt{t^2+1}\right|\right]+C \\ =\frac{1}{2}\left[\frac{x^2}{2}\sqrt{x^4+1}+\frac{1}{2}\log \left|x^2+\sqrt{x^4+1}\right|\right]+C \\ =\frac{x^2}{4}\sqrt{x^4+1}+\frac{1}{4}\log \left|x^2+\sqrt{x^4+1}\right|+C \end{gathered}$$