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...
 A: 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.
A: An Euler substitution 
$$t=\sqrt{x^2+1}-x$$
will reduce the integral to a rational function.
A: You have to be a little hyperbolic savy to do this method but $x=\sinh u\to \text{d}x=\cosh u\text{d}u$ so the integral is now $\int \sinh^2 u\cosh^2 u\text{d}u=\int\frac{\sinh^2 2u}{4}\text{d}u=\int\frac{\cosh 4u-1}{8}\text{d}u=\frac{\frac{\sinh 4u}{4}-u}{8}=\frac{\sinh u \cosh u\cosh 2u-u}{8}=\frac{x\sqrt{1+x^2}\left ( 1+2x^2\right )-\sinh^{-1} x}{8}+C=\frac{x\sqrt{1+x^2}\left ( 1+2x^2\right )-\ln \left (x+\sqrt{1+x^2}\right )}{8}+C$
A: Try the following substitution
$$x=\frac{e^t-e^{-t}}{2}$$
and you'll get an easy integral.
A: $$\int\frac{\sin^2 t}{\cos^7 t}\;dt=\int\frac{\sin^2 t \cos t}{(1-\sin^2 t)^4}\;dt= \int \frac{u^2}{(1-u^2)^4} du \\
= \int \frac{u^2}{(1-u)^4(1+u)^4} du$$
and partial fractions.
An alternate way is to observe that 
$$\int\frac{\tan^2 t}{\cos^5 t}\;dt= \int \tan^2 t \sec^5 t dt$$
and use the reduction formula .
