Evaluate $\int_0^\infty\frac{dl}{(r^2+l^2)^{\frac32}}$ How to evaluate the following integral
$$\int_0^\infty\frac{dl}{(r^2+l^2)^{\large\frac32}}$$
The solution is supposed to look like this, unfortunately I can't derive it.
$$
\left[\frac{l}{r^2\sqrt{r^2+l^2}}\right]_{l=0}^\infty
$$
 A: Let $l=r\tan{u}$, then $dl=r\sec^2{u} \ du$. The integral becomes
$$\frac{1}{r^2}\int^{\pi/2}_0\frac{\sec^2{u}}{\sec^3{u}}du=\frac{1}{r^2}$$
A: All integrals of the form $\displaystyle\int_0^\infty\frac{x^{k-1}}{(x^n+a^n)^m}dx$ can be expressed in terms of trigonometric functions 
as follows: first, let $x=at$, then $u=\dfrac1{t^n+1}$. Now recognize the expression of the beta function in 
the new integral, and finally use Euler's reflection formula for the $\Gamma$ function in order to arrive at 
the desired result.
A: $$
\int\frac{dl}{\left(r^2+l^2\right)^{3/2}} = -\frac{1}{r}\frac{d}{dr}\int\frac{1}{\left(r^2+l^2\right)^{1/2}}dl
$$
change of variables $v=l/r$
we find
$$
-\frac{1}{r}\frac{d}{dr}\frac{1}{r}\int^{\infty}_{0}\frac{1}{\left(1+v^2\right)^{1/2}}rdv = -\frac{1}{r}\frac{d}{dr}\left[\sinh^{-1}v\right]^{\infty}_{0}
$$
$$
-\frac{1}{r}\frac{d}{dr}\left[\sinh^{-1}v\right]^{\infty}_{0} = -\frac{1}{r}\left[\frac{1}{\sqrt{1+v^2}}\frac{dv}{dr}\right]^{\infty}_{0} = -\frac{1}{r}\left[\frac{1}{\sqrt{1+\left(\frac{l}{r}\right)^2}}\left(\frac{-l}{r^2}\right)\right]^{\infty}_{0} 
$$
which leads to your result.
A: $$I=\int_0^\infty \frac{dx}{(x^2+r^2)^{3/2}}$$
Let $xr=\tan u$. Therefore $dx=\frac1r\sec^2u\ du$:
$$I=\frac1r\int_0^{\pi/2}\frac{\sec^2u\ du}{(r^2\tan^2u+r^2)^{3/2}}$$
$$I=\frac1{r^2}\int_0^{\pi/2}\frac{\sec^2u\ du}{(\tan^2u+1)^{3/2}}$$
$$I=\frac1{r^2}\int_0^{\pi/2}\frac{\sec^2u\ du}{(\sec^2u)^{3/2}}$$
$$I=\frac1{r^2}\int_0^{\pi/2}\frac{\sec^2u\ du}{\sec^3u}$$
$$I=\frac1{r^2}\int_0^{\pi/2}\frac{du}{\sec u}$$
$$I=\frac1{r^2}\int_0^{\pi/2}\cos u\ du$$
$$I=\frac1{r^2}$$
