Integrate $\int{ \frac{z - Ru}{(R^2  + z^2 - 2Rzu)^{3/2}} du }$ This integral comes from a physics book when calculating a field of an uniformly charged sphere (without Gauss' Law).
It says that it can be done by partial fractions, but I cannot imagine how.
 A: As GEdgar pointed out in a comment, this is
$$\frac{1}{2}\int\frac{f'(z)}{f(z)^{3/2}}\mathrm du\;,$$
with $f(z)=R^2  + z^2 - 2Rzu$, so perhaps the easiest way to do this is
$$
\begin{eqnarray}
\frac{1}{2}\int\frac{f'(z)}{f(z)^{3/2}}\mathrm du
&=&
-\int\frac{\mathrm d}{\mathrm dz}f(z)^{-1/2}\mathrm du
\\
&=&
-\frac{\mathrm d}{\mathrm dz}\int f(z)^{-1/2}\mathrm du
\\
&=&
-\frac{\mathrm d}{\mathrm dz}\int(R^2  + z^2 - 2Rzu)^{-1/2}\mathrm du
\\
&=&
\frac{\mathrm d}{\mathrm dz}\frac{(R^2  + z^2 - 2Rzu)^{1/2}}{Rz}\;.
\end{eqnarray}
$$
A: If it is indeed the integral with respect to $u$ that you want, and $z$ is an oddly named constant, make the substitution $w=R^2+z^2 -2Rzu$.  Despite appearances, this is a simple linear substitution. 
Go through the process, not forgetting that $dw=-2Rz du$.
You end up with an integral of the shape
$$\int \frac{A+Bw}{w^{3/2}} dw.$$
where $A$ and $B$ are somewhat messy constants.
Split the integrand into two parts, $A/w^{3/2}$ and $B/w^{1/2}$.
If in your integral $du$ is a typo for $dz$, life is even simpler, make the same substitution, and you end up with $dw/2$ on top.
A: Let $\sqrt{R^2+z^2-2Rzu}=t$ and then differentiating both sides gives
$\dfrac{-2Rz}{2\sqrt{R^2+z^2-2Rzu}} du = dt$, which is equivalent to $\dfrac{-Rz}{t}du=dt$. Substituting this, 
$\displaystyle \int{ \frac{z-\frac{R^2+z^2-t^2}{2z}}{t^3} \frac{-t}{Rz} dt}=\int{ \frac{-R^2+z^2+t^2}{2z} \frac{-1}{Rzt^2} dt} = \frac{1}{2Rz^2} \int{ (\frac{R^2-z^2}{t^2} -1) dt}$.
I guess you can do it now.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large%
\int_{-1}^{1}{z - Ru \over \pars{R^{2}  + z^{2} - 2Rzu}^{3/2}}\,\dd u}
=\int_{u\ =\ -1}^{u\ =\ 1}\pars{z - Ru}\,
\dd\pars{1/\pars{Rz} \over \root{R^{2}  + z^{2} - 2Rzu}}
\\[5mm]&=\left.{1 \over Rz}\,
{z - Ru \over \root{R^{2}  + z^{2} - 2Rzu}}\,\right\vert_{\,u\ =\ -1}^{\,u\ =\ 1}
-\int_{-1}^{1}{1 \over Rz}\,{-R \over \root{R^{2}  + z^{2} - 2Rzu}}\,\dd u
\\[1cm]&={1 \over Rz}\pars{{z - R \over \root{R^{2} + z^{2} - 2Rz}}
-{z + R \over \root{R^{2} + z^{2} + 2Rz}}}
\\[5mm]&\phantom{=}+{1 \over z\,d_{>}}
\int_{-1}^{1}{\dd u \over \root{1 - 2\pars{d_{<}/d_{>}}u + \pars{d_{<}/d_{>}}^{2}}}
\quad\mbox{where}\quad
\left\{\begin{array}{lcl}
d_{<} & \equiv & \min\pars{\verts{R},\verts{z}}
\\[2mm]
d_{>} & \equiv & \max\pars{\verts{R},\verts{z}}
\end{array}\right.
\end{align}

In the last integral, the integrand can be expanded in Legendre Polynomias
  $\ds{\,{\rm P}_{\ell}:\bracks{-1,1} \to {\mathbb R}}$, with
  $\ds{\ell = 0,1,2,3,\ldots}$, which satisfy
  $\ds{\int_{-1}^{1}\,{\rm P}\ell\pars{x}\,{\rm P}\ell'\pars{x}
     ={2\,\delta_{\ell\ell'} \over 2\ell + 1}}$. Then,

\begin{align}&\color{#66f}{\large%
\int_{-1}^{1}{z - Ru \over \pars{R^{2}  + z^{2} - 2Rzu}^{3/2}}\,\dd u}
\quad\pars{~\mbox{Note that}\ \,{\rm P}_{0}\pars{x} = 1\,,\
\forall\ x \in \bracks{-1,1}~}
\\[5mm]&={\sgn\pars{z - R} - \sgn\pars{z + R} \over Rz}
+{1 \over z\,d_{>}}
\sum_{\ell\ =\ 0}^{\infty}\pars{d_{<} \over d_{>}}^{\ell}\ \overbrace{%
\int_{-1}^{1}\,{\rm P}_{\ell}\pars{u}\,\dd u}
^{\ds{=\ \dsc{2\,\delta_{\ell,0}}}}
\\[5mm]&=\color{#66f}{\large%
{1 \over z}\bracks{{\sgn\pars{z - R} - \sgn\pars{z + R} \over R} +{2 \over d_{>}}}}
\end{align}

where
  $\ds{\quad\left\{\begin{array}{lcl}
d_{<} & \equiv & \min\pars{\verts{R},\verts{z}}
\\[2mm]
d_{>} & \equiv & \max\pars{\verts{R},\verts{z}}
\end{array}\right.}$

