How to evaluate integral: $\int x^2\sqrt{x^2+1}\;dx.$ I want to solve the integral:$$\int x^2\sqrt{x^2+1}\;dx.$$
I did $x = \tan t$, then it is equal to:$$\int\frac{\tan^2 t}{\cos^3 t}\;dx.$$
Or:$$\int\frac{\sin^2 t}{\cos^5 t}\;dx.$$
I stuck there. Any help will be much appreciated...

Going the hyperbolic instead of the trigonometric route, we can let $x:=\sinh t$, then $dx=\cosh tdt$.
\begin{align}
\int x^2\sqrt{x^2+1}dx &= \int\sinh^2t\sqrt{\sinh^2t+1}\cosh tdt\\
&=\int\sinh^2t\cosh^2tdt\\
&=\int\left(\sinh t\cosh t\right)^2dt\\
&=\frac{1}{4}\int\sinh^2(2t)dt\\
&=\frac{1}{8}\int\left(\cosh(4t)-1\right)dt\\
&=\frac{1}{32}\left(\sinh(4t)-4t\right) + C
\end{align}
Since $t=\text{arsinh }x$ by inversion of our initial substitution, we have
\begin{align}
\int x^2\sqrt{x^2+1}dx &= \frac{1}{32}\left[\sinh(4\text{arsinh }x)-4\text{arsinh }x\right]+C\\
&=\frac{1}{8}\left((x+2x^3)\sqrt{x^2+1}-\text{arsinh }x\right) + C
\end{align}
If you want, you can write out $\text{arsinh }x$ using logarithms.
