Integrating $x^3 \sqrt{x^2+1} $

Question

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?

Answer 1

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 $u\,du$.

Answer 2

By substituting $x=\sinh t$ we have: $$ I = \int \sinh^3 t\cosh^2 t\,dt =\int \sinh^3 t\,dt + \int\sinh^5 t\,dt.$$ Just turn $\sinh t$ into $\frac{e^t-e^{-t}}{2}$ and integrate termwise what you get through the binomial theorem.