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.
$$
\left[\frac{l}{r^2\sqrt{r^2+l^2}}\right]_{l=0}^\infty
$$

Let $l=r\tan{u}$, then $dl=r\sec^2{u} \ du$. The integral becomes
$$\frac{1}{r^2}\int^{\pi/2}_0\frac{\sec^2{u}}{\sec^3{u}}du=\frac{1}{r^2}$$
