Integral $\int^\ell_{-\ell} r/\sqrt{L^2+r^2}^{\,3} \, dL$ We tried to solve a magnetics problem and ended up with 
$$\int^\ell_{-\ell} \frac{r}{\sqrt{L^2+r^2}^{\,3}} \, dL.$$
How do I solve this integral?
 A: Set $L=rx$, so that $dL = r\cdot dx$ and the given integral becomes:
$$ \frac{1}{r}\int_{-l/r}^{l/r}\frac{dx}{\left(\sqrt{x^2+1}\right)^3}=\frac{2}{r}\cdot\left. \frac{x}{\sqrt{1+x^2}}\,\right|_{0}^{l/r}=\color{red}{\frac{2 l}{r\sqrt{r^2+l^2}}}. $$
A: $$\int_{-l}^{l}\frac{r}{\left(\sqrt{L^2+r^2}\right)^3}\space\text{d}L=$$
$$r\int_{-l}^{l}\frac{1}{\left(\sqrt{L^2+r^2}\right)^3}\space\text{d}L=$$
$$r\int_{-l}^{l}\frac{1}{\left(L^2+r^2\right)^{\frac{3}{2}}}\space\text{d}L=$$

For the integrand $\frac{1}{\left(L^2+r^2\right)^{\frac{3}{2}}}$, (assuming all variables are positive).
Substitute $L=r\tan(u)$ and $\text{d}L=r\sec^2(u)\space\text{d}u$.
Then $\left(L^2+r^2\right)^{\frac{3}{2}}=\left(r^2\tan^2(u)+r^2\right)^{\frac{3}{2}}=r^3\sec^3(u)$ and $u=\arctan\left(\frac{L}{r}\right)$.
This gives a new lower bound $u=-\arctan\left(\frac{l}{r}\right)$ and upper bound $u=\arctan\left(\frac{l}{r}\right)$:

$$r^2\int_{-\arctan\left(\frac{l}{r}\right)}^{\arctan\left(\frac{l}{r}\right)}\frac{\cos(u)}{r^3}\space\text{d}u=$$
$$\frac{1}{r}\int_{-\arctan\left(\frac{l}{r}\right)}^{\arctan\left(\frac{l}{r}\right)}\cos(u)\space\text{d}u=$$
$$\frac{1}{r}\left[\sin(u)\right]_{-\arctan\left(\frac{l}{r}\right)}^{\arctan\left(\frac{l}{r}\right)}=$$
$$\frac{1}{r}\left(\sin\left(\arctan\left(\frac{l}{r}\right)\right)-\sin\left(-\arctan\left(\frac{l}{r}\right)\right)\right)=$$
$$\frac{1}{r}\left(\frac{l}{\sqrt{l^2+r^2}}--\frac{l}{\sqrt{l^2+r^2}}\right)=$$
$$\frac{1}{r}\left(\frac{l}{\sqrt{l^2+r^2}}+\frac{l}{\sqrt{l^2+r^2}}\right)=$$
$$\frac{1}{r}\left(\frac{2l}{\sqrt{l^2+r^2}}\right)=$$
$$\frac{\frac{2l}{\sqrt{l^2+r^2}}}{r}=$$
$$\frac{2l}{r\sqrt{l^2+r^2}}$$
