<s>[INST] 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.
 [/INST] 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 $3^x-1=v$
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