Can anyone tell me if I made any errors in my solution and whether or not my answer is correct? Is there a better/quicker method to solving this?
If you can't see the writing very well (and therefore can't see my work and answer very well), then note the problem is read as "the integral from 0 to infinity of $\cos(bx)(x-\log(e^x-1)~dx$ where $b>0$"
\begin{align} \require{cancel} \int_0^\infty e^{-ax}\cos(bx)~dx&=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}\int_0^\infty\cos(bx)(bx)^{2n}~dx\\&=\sum_{n=0}^\infty\frac{(-b^2)^n}{(2n)!}\int_0^\infty e^{-ax}x^{2n}~dx \quad\boxed{u=ax,~du=a~dx,~x=\frac ua} \\ &= \frac1a\sum_{n=0}^\infty\frac{(-b^2)^n}{(2n)!(a^2)^n}\int_0^\infty u^{2n}e^{-u}~du\\&= \frac1a\sum_{n=0}^\infty\left(\frac{-b^2}{a^2}\right)^n\frac{\cancel{(2n)!}}{\cancel{(2n)!}}\\&= \frac1a\sum_{n=0}^\infty\left(\frac{-b^2}{a^2}\right)^n=\frac1a\left[\frac1{1+\frac{b^2}{a^2}}\right]=\frac a{a^2+b^2} \end{align}
\begin{align} \require{cancel} \int_0^\infty e^{-ax}\cos(bx)~dx=\frac a{a^2+b^2}\implies \frac1{\cancel a}\left[\frac{\cancel a}{a^2+b^2}\right]&=\frac1a\int_0^\infty e^{-ax}\cos(bx)~dx\\&= \sum_{a=1}^\infty\frac1{a^2+b^2}=\sum_{a=1}^\infty\int_0^\infty\frac{e^{-ax}}a~dx \end{align}
$$\implies \frac{\pi b\coth(\pi b)-1}{2b^2}=\int_0^\infty\cos(bx)\sum_{a=1}^\infty\frac{e^{-ax}}a~dx\implies \sum_{a=1}^\infty\frac{e^{-ax}}a=\frac{e^{-x}}{1-e^{-x}}=\frac1{e^x-1}$$
\begin{align} \implies-\sum_{a=1}^\infty\frac{e^{-ax}}a=\int\frac1{e^x-1}~dx&\implies\sum_{a=1}^\infty\frac{e^{-ax}}a=-(\ln(e^x-1)-x)=x-\ln(e^x-1)\end{align}
$$\implies\small\sum_{a=1}^\infty\frac{e^{-ax}}a=\int_0^\infty\cos(bx)\sum_{a=1}^\infty\frac{e^{-ax}}a~dx=\int_0^\infty\cos(bx)(x-\ln(e^x-1))~dx=\frac{\pi b\coth(\pi b)-1}{2b^2}$$
You want to double-check your correct conclusion that if $b>0$ then $$\int_0^\infty\cos bx \ln(1-e^{-x})dx=\frac{1-\pi b \coth \pi b}{2b^2}.$$Here's a slicker way: $$-\sum_{n\ge 1}\frac{1}{n}\Re \int_0^\infty\exp-(n-ib)xdx=-\sum_{n\ge 1}\frac{1}{n^2+b^2}.$$Finish with $$\coth \pi b=\frac{1}{\pi b}+\frac{2b}{\pi}\sum_n\frac{1}{n^2+b^2}.$$