Show that $\int_0^\pi\frac{x \sin x}{5-3\cos x}\, \mathrm dx = \frac{2 \pi \log (4/3)} 3$ 
Show that $$\int_0^\pi\frac{x \sin x}{5-3\cos x}\, \mathrm dx = \frac{2 \pi \log (4/3)} 3$$

I am struggling with this one. I have tried substitutions and the residue theorem but haven't got anywhere. It would be nice to see an 'elementary' solution if there is one, but any will do.
 A: There's a simple trick:
$$\sum_{n\geq 1}\frac{\sin(nx)}{3^n} = \frac{3}{2}\cdot\frac{\sin x}{5-3\cos x}\tag{1} $$
and
$$ \int_{0}^{\pi} x \sin(nx)\,dx = \frac{\pi(-1)^{n+1}}{n}\tag{2}$$
hence
$$ \int_{0}^{\pi}\frac{x\sin(x)}{5-3\cos x}\,dx = \frac{2\pi}{3}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n 3^n}=\frac{2\pi}{3}\log\left(1+\frac{1}{3}\right) \tag{3}$$
nice & easy.
A: First integrate by parts to get:
\begin{align}
\int^\pi_0\frac{x\sin(x)}{5-3\cos(x)}\,dx= \frac{\pi}{3}\log(8)-\frac{1}{3}\int^\pi_0 \log(5-3\cos(x))\,dx
\end{align}
The integral on the right, can be done with many methods, for example, notice that we have with substituting $u=\pi-x$ that: 
\begin{align}
\int^\pi_0 \log(5-3\cos(x))\,dx&=\frac{1}{2}\int^\pi_0 \log(25-9\cos^2(x))\,dx\\
&=\frac{\pi}{2}\log(25)+\frac{1}{2}\int^\pi_0\log\left(1-\frac{9}{25}\cos^2(x)\right)\,dx\\
&=\frac{\pi}{2}\log(25)+\int^{\pi/2}_0\log\left(1-\frac{9}{25}\sin^2(x)\right)\,dx\\
\end{align}
We know from this link that: $$\int^{\pi/2}_0\log\left(1-\frac{9}{25}\sin^2(x)\right)\,dx=\pi\log(9/10)$$
Putting everything together yields: 
\begin{align}
\int^\pi_0\frac{x\sin(x)}{5-3\cos(x)}\,dx=\frac{\pi}{3}\log(8)-\frac{1}{3}\left( \frac{\pi}{2}\log(25)+\pi\log(9/10)\right)=\frac{2\pi\log(4/3)}{3}\end{align}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
&\bbox[10px,#ffd]{\ds{%
\int_{0}^{\pi}{x\sin\pars{x} \over 5 - 3\cos\pars{x}}\,\dd x}} =
\left.\Re\int_{x\ =\ 0}^{x\ =\ \pi}{\bracks{-\ic\ln\pars{z}}\bracks{\pars{1 - z^{2}}\ic/\pars{2z}} \over 5 - 3\bracks{\pars{1 + z^{2}}/\pars{2z}}}\,
{\dd z \over \ic z}\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\\[5mm] = &\
\left.-\,{1 \over 3}\,\Im\int_{x\ =\ 0}^{x\ =\ \pi}\ln\pars{z}\,
{\pars{1 - z^{2}} \over \pars{z - 1/3}\pars{z - 3}}\,
{\dd z \over z}\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\qquad\pars{\begin{array}{l}
\mbox{with}
\\
\ds{-\,{\pi \over 2} < \arg\pars{z} < {3\pi \over 2}}
\end{array}}
\\[1cm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\, &\
{1 \over 3}\,\Im\int_{-1}^{-\epsilon}\bracks{\ln\pars{-x} + \ic\pi}\,
{\pars{1 - x^{2}} \over \pars{x - 1/3}\pars{x - 3}}\,{\dd x \over x}
\\[2mm] + &\
{1 \over 3}\,\Im\int_{\pi}^{0}\bracks{\ln\pars{\epsilon} + \ic\theta}\
\,{\epsilon\expo{\ic\theta}\ic\,\dd\theta \over \epsilon\expo{\ic\theta}} +
{1 \over 3}\,\Im\int_{\epsilon}^{1/3-\epsilon}
\ln\pars{x}\,{\pars{1 - x^{2}} \over \pars{x - 1/3}\pars{x - 3}}\,
{\dd x \over x}
\\[2mm] + &\
{1 \over 3}\,\Im\int_{\pi}^{0}\ln\pars{1 \over 3}\
{8/9 \over \pars{\epsilon\expo{\ic\theta}}\pars{-8/3}}\,{\epsilon\expo{\ic\theta}\ic\,\dd\theta \over 1/3} +
{1 \over 3}\,\Im\int_{1/3 + \epsilon}^{1}
\ln\pars{x}\,{\pars{1 - x^{2}} \over \pars{x - 1/3}\pars{x - 3}}\,
{\dd x \over x}
\\[1cm] = &\
-\,{1 \over 3}\,\pi\int_{\epsilon}^{1}{1 - x^{2} \over \pars{x + 1/3}
\pars{x + 3}}\,{\dd x \over x} -
{1 \over 3}\,\pi\ln\pars{\epsilon} - {1 \over 3}\,\pi\ln\pars{3}
\\[5mm] =  &\
-\,{1 \over 3}\,\pi\int_{\epsilon}^{1}\bracks{{1 - x^{2} \over \pars{x + 1/3}
\pars{x + 3}} - 1}\,{\dd x \over x}\ \underbrace{-
{1 \over 3}\,\pi\int_{\epsilon}^{1}{\dd x \over x} -
{1 \over 3}\,\pi\ln\pars{\epsilon}}_{\ds{=\ 0}}\ - {1 \over 3}\,\pi\ln\pars{3}
\\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\to}\,\,\, &\
\pi\
\underbrace{\int_{0}^{1}\pars{{1 \over 1 + 3x} + {1 \over 9 + 3x}}\dd x}
_{\ds{{1 \over 3}\,\ln\pars{16 \over 3}}} -
{1 \over 3}\,\pi\ln\pars{3} = \bbx{{2 \over 3}\,\pi\ln\pars{4 \over 3}}
\end{align}
