Integrating $x^3 \sqrt{x^2+1} $ I need to solve this integral $\int x^3 \sqrt{x^2+1} dx$
Someone could explain to me how?
I have tried to use substitutions, i.e. $x^2+1$ and  $\sqrt{x^2+1}$; but seems like its not correct. Someone could explain to me how?

The substitution $u=\sqrt{x^2+1}$, or equivalently $u^2=x^2+1$, works nicely. We get $u\,du=x\,dx$. Borrow an $x$ from $x^3$ to keep $dx$ company. 
We want 
$$\int x^2\sqrt{x^2+1}\,\, x\,dx.$$
The term $x^2$ is $u^2-1$. The term $\sqrt{x^2+1}$ is $u$. And finally the $x\,dx$ part is $u\,du$.  
