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.

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}
