Integration By Parts Problem I am stuck on the following problem.
$$\int\frac{r^3}{\sqrt{4+r^2}} dr$$ 
Using Integreation by parts with $u = r^2$
I am not allowed to use a different $u$.
This is my work so far:
$$u = r^2 \rightarrow du = 2rdr$$ 
$$dv = (4+u^2)^{-1/2} \rightarrow v = ?$$
I am not sure where to go from here. Any help would be appreciated.
 A: If you're using $u=r^2$, then it looks like we should have $dv=\frac{r\,dr}{\sqrt{4+r^2}}$. That can be integrated by ordinary substitution, and we see that $v=\sqrt{4+r^2}$. Now you can apply the integration by parts formula to obtain
$\displaystyle{\int} r^2 \frac{r\,dr}{\sqrt{4+r^2}} = r^2\sqrt{4+r^2} - \displaystyle{\int} 2r\sqrt{4+r^2}dr$
This second integral can be completed by ordinary substitution, too.
Does that give you a push in the right direction?
A: Let $u=r^2$. Then as you have done, we have $du=2rdr$. Hence, we have 
$$\int\frac{r^3}{\sqrt{4+r^2}} dr=\frac{1}{2}\int\frac{r^2 \cdot 2rdr}{\sqrt{4+r^2}} 
=\frac{1}{2}\int\frac{u du}{\sqrt{4+u}}.$$
I leave it to you from here. 
A: without integration by parts : set $r^2+4=u^2$ then $2rdr=2udu$ and $r^2=u^2-4$ so it follows that $$\int \frac{r^2}{\sqrt{r^2+4}}dr=\int\frac{(u^2-4)udu}{u}=\frac{u^3}{3}-4u+c.$$
so changing the variable u with $\sqrt{r^2+4}$ you obtain $$\frac{(r^2+4)^{3/2}}{3}-4\sqrt{r^2+4}+c.$$
A: Let $r=2\sinh x$ then $dr=2\cosh x \,dx$ and $\sqrt{4+r^2}=2\cosh x$ so
$$\int\frac{r^3}{\sqrt{4+r^2}} dr=8\int\sinh^3 x\,dx=8\int\cosh^2x\sinh x\,dx-8\int\sinh x\,dx\\=\frac 8 3\cosh^3 x-8\cosh x+C$$ 
