Help with closed form of $\int_0^\infty\frac{\tanh(ax)}{e^x-1}dx$ I have been trying to find the value of:$$\int_0^\infty\frac{\tanh(ax)}{e^x-1}dx=\int_0^\infty\frac{e^{2ax}-1}{(e^{2ax}+1)(e^x-1)}dx$$
Under u-substitution:
Let $u=e^{ax}$, $x=\frac{\log(u)}{a}$, $dx=\frac{du}{ua}$
$$\int_1^\infty \frac{u^2-1}{(u^2+1)(u^{\frac{1}{a}}-1)}\frac{du}{ua}$$
I don't really know where to proceed further from here or if I should use a series expansion of $\tanh(ax)$.

Not an answer.
Since marty cohen started with some values, I shall add a few other (including those already given)
$$\left(
\begin{array}{cc}
a & \text{result} \\
 \frac{1}{2} & \log (2) \\
 1 & \frac{1}{4} (\pi +2\log (2)) \\
 \frac{3}{2} & \frac{1}{9} \left(2 \sqrt{3} \pi +3\log (2)\right) \\
 2 & \frac{1}{8} \left(\pi +2 \sqrt{2} \pi +2\log (2)\right) \\
 3 & \frac{1}{36} \left(\left(15+4 \sqrt{3}\right) \pi +6\log (2)\right) \\
 4 & \frac{1}{16} \left(\pi +2 \sqrt{2} \pi +8 \pi  \cos \left(\frac{\pi
   }{8}\right)+2\log (2)\right)
\end{array}
\right)$$ Using numerical integration (let $a=10^k$), we have as results
$$\left(
\begin{array}{cc}
 k & \text{integral} \\
 0 & 1.13197 \\
 1 & 3.15568 \\
 2 & 5.42741 \\
 3 & 7.72688 \\
 4 & 10.0292 \\
 5 & 12.3317 \\
 6 & 14.6343 \\
 7 & 16.9369 \\
 8 & 19.2395 \\
 9 & 21.5420 \\
 10 & 23.8446
\end{array}
\right)$$ which seems to be almost a straight line when plotted as a function of $k$ $(R^2=0.99997)$ and the slope seems to be very close to $\log(10)$.
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
 a & 0.928645 & 0.049392 & \{0.814746,1.042544\} \\
 b & 2.287000 & 0.008349 & \{2.267748,2.306253\} \\
\end{array}$$
