Finding $\int_{1}^{\infty} \frac{x 3^x}{(3^x-1)^2}$ How do we prove that $$\int_{1}^{\infty} \frac{x \cdot 3^x}{(3^x-1)^2} \mathrm{d}x=\dfrac{3\log{3}-2\log{2}}{2\log^2{3}}$$
I have tried some substitutions such as $3^x=t$, but it didn't work out. The denominator is already factored. I do not know what to do next. Please help me out. Thank you.
 A: Integrating by parts ($u=x$, $dv=\frac{3^x}{(3^x-1)^2}dx$), $$\int x\frac{3^x}{(3^x-1)^2}dx=x\int\frac{3^x}{(3^x-1)^2}dx-\int\left[\frac{dx}{dx}\int\frac{3^x}{(3^x-1)^2}dx\right]dx$$
Now for $\displaystyle\int\dfrac{3^x}{(3^x-1)^2}dx,$  set $3^x-1=u$
Finally $\displaystyle\int\dfrac{dx}{3^x-1},$ set $3^x-1=v$
A: Another answers are nice but if you want to stick to Integration by parts, Here's how
$$I_1=\int_1^\infty\frac{x 3^x}{(3^x-1)^2}$$
$$I=\int\frac{x 3^x}{(3^x-1)^2}$$
substitute
$$3^x-1=t\iff3^x\ln 3=dt$$
$$I=\int\frac{x \cdot 3^x}{(3^x-1)^2} \mathrm{d}x= \frac{1}{\ln^23}\int \frac{\ln|t+1|}{t^2} \mathrm{d}t$$
$$I(\ln^23)= \int \frac{\ln|t+1|}{t^2} \mathrm{d}t=-\frac{\ln|t+1|}{t}+\int \frac{1}{t(t+1)} \mathrm{d}t$$
$$I(\ln^23)=-\frac{\ln|t+1|}{t}-\ln|t|+\ln|t+1|$$
$$I=\frac{(3^x-1)\ln|1-3^x|-3^xx\ln(3)}{(\ln^23)(3^x-1)}$$
$$\rightarrow I_1=\frac{\ln27-\ln4}{2\ln^23}$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\int_{1}^{\infty}{x\ 3^{x} \over \pars{3^{x} - 1}^{2}}\,\dd x
     ={3\ln\pars{3} - 2\ln\pars{2} \over 2\ln^{2}\pars{3}}:\ {\large ?}}$.

\begin{align}&\color{#c00000}{\int_{1}^{\infty}{\dd x \over \expo{\mu x} - 1}}
=\int_{1}^{\infty}{\expo{-\mu x}\,\dd x \over 1 - \expo{-\mu x}}
=\left.{\ln\pars{1 - \expo{-\mu x}} \over \mu}
\right\vert_{x\ =\ 1}^{x\ \to\ \infty}
=-\,{\ln\pars{1 - \expo{-\mu}} \over \mu}
\end{align}

Derivate both members respect of $\ds{\mu}$:

\begin{align}&\color{#c00000}{\int_{1}^{\infty}%
\bracks{-\,{x\expo{\mu x} \over \pars{\expo{\mu x} - 1}^{2}}}\,\dd x}
=-\,{1 \over \mu\pars{\expo{\mu} - 1}} + {\ln\pars{1 - \expo{-\mu}} \over \mu^{2}}
\end{align}

Multiply both members by $\ds{-1}$ and set $\ds{\mu = \ln\pars{3}}$:

\begin{align}&\color{#66f}{\large\int_{1}^{\infty}%
{x\ 3^{x} \over \pars{3^{x} - 1}^{2}}\,\dd x}
={1 \over \ln\pars{3}\pars{3 - 1}} - {\ln\pars{1 - 1/3} \over \ln^{2}\pars{3}}
=\color{#66f}{\large{3\ln\pars{3} - 2\ln\pars{2} \over 2\ln^{2}\pars{3}}}
\\[5mm]&\approx {\tt 0.7911}
\end{align}

