Hint for solving the following integral: $$\int \frac{2r}{|1-r^{2}|^{\frac{3}{2}}}dr?$$
My attempt: I tried putting $u= \frac{1}{\sqrt{|1-r^{2}|}}$ then $du = \frac{r}{(1-r^{2})\sqrt{|1-r^{2}|}}$, but not working. Can someone help me in solving this question?
I would be happy to get some hints.
Let $I$ denote your integral. Note that $1-r^2=0$ for $r=\pm1$, such that we distinguish two cases: $r\in(-1,1)$ and $r\notin(1,1)$.
$r\in(1,1)$. Then, with $u=1-r^2$ and thus $du=-2rdr$, $$I=\int{\frac{2r}{(1-r^2)^{\frac{3}{2}}}dr}=-\int{\frac{1}{u^{\frac{3}{2}}}du}=-\int{u^{-\frac{3}{2}}du}=\frac{2}{\sqrt{u}}=\frac{2}{\sqrt{1-r^2}}$$
$r\notin(1,1)$. Then, with $u=r^2-1$ and thus $du=2rdr$, $$I=\int{\frac{2r}{(r^2-1)^{\frac{3}{2}}}dr}=\int{\frac{1}{u^{\frac{3}{2}}}du}=\int{u^{-\frac{3}{2}}du}=-\frac{2}{\sqrt{u}}=-\frac{2}{\sqrt{1-r^2}}$$
Concluding: $$\int{\frac{2r}{\left|1-r^2\right|^{\frac{3}{2}}}dr}=\begin{cases} \frac{2}{\sqrt{1-r^2}} & \mbox{ if } x\in(1,1) \\ -\frac{2}{\sqrt{1-r^2}} & \mbox{ if } x\notin(1,1) \end{cases}$$