$$ I=\int_0^\infty \frac{x \cdot e^{-2x}}{\sqrt{e^x-1}} \, dx $$
Let $(u=\sqrt{e^x-1})$ which gives $(\ln(u^2+1)=x) $ and $(\frac{2u \, du}{u^2+1}= dx)$.
The new bounds of integration are still 0 and infinity.
Now we have
$$ \int_{0}^{\infty} \frac{\ln(u^{2}+1)}{\left(e^{2ln(u^{2}+1)}\right) \cdot u} \cdot \frac{2u}{u^{2}+1} \, du $$
Bounds: From 0 to infinity.
After simplification using log rules, we have $(\frac{\ln(u^2+1)}{(u^2+1)^3})$ and the $(u)$ cancels, leaving us with
$ \int_0^\infty \frac{\ln(u^2+1)}{(u^2+1)^3} \, du \quad \text{or} \quad \int_0^\infty 2 \ln(u^2+1) \cdot (u^2+1)^{-3} \, du $
Then using $(u=\tan(w))$, $(du=\sec(w)^2)$, and
$ ln(u^2+1) \rightarrow \ln(\tan(w)^2+1) \rightarrow \ln(1/\sec(w)^2) \rightarrow -2 \ln(\cos(w)) $
we are left with
$ \int_0^{\frac{\pi}{2}} -4 \ln(\cos(w)) \cdot \cos(w)^4 \, dw $
Would using complex analysis be appropriate to solve? If so, how would I evaluate the closed form of this integral that I left off on? If not, how do I consider the integral from where I left off? Thanks
NOTE: The original post asked to evaluate the integral $\int_0^\infty \frac{x^2e^x}{(e^x-1)^{3/2}}\,dx$. The following development began before any revisions. However,the methodology used here can be applied to the integral in the revised post (That solution is presented in the BONUS section).
Note that we have
$$\begin{align} \int_0^\infty \frac{x^2e^x}{(e^x-1)^{3/2}}\,dx&=\int_1^\infty \frac{\log^2(x)}{(x-1)^{3/2}}\,dx\tag1\\\\ &=\int_0^\infty \frac{\log^2(x+1)}{x^{3/2}}\,dx\tag2\\\\ &=4\int_0^\infty \frac{\log(x+1)}{x^{1/2}(x+1)}\,dx\tag3\\\\ &=4\frac{d}{dt}\left.\left(\int_0^\infty \frac{(x+1)^t}{x^{1/2}}\,dx\right)\right|_{t=-1}\tag4\\\\ &=4\frac{d}{dt}\left.\left(2\int_0^\infty (x^2+1)^t\,dx\right)\right|_{t=-1}\tag5\\\\ &=4\frac{d}{dt}\left.\left(2\int_0^{\pi/2} \sec^{2t+2}(x)\,dx\right)\right|_{t=-1}\tag6\\\\ &=4\frac{d}{dt}\left.\left(B(1/2,-t-1/2)\right)\right|_{t=-1}\tag7\\\\ &=4\frac{d}{dt}\left.\left(\frac{\Gamma(1/2)\Gamma(-t-1/2)}{\Gamma(-t)}\right)\right|_{t=-1}\tag8\\\\ &=4\sqrt\pi\left.\left(-\frac{\Gamma(-t-1/2)\psi(-t-1/2)}{\Gamma(-t)}\right)\right|_{t=-1}\\\\ &+4\sqrt\pi\left.\left(\frac{\Gamma(-t)\Gamma(-t-1/2)\psi(-t)}{\Gamma^2(-t)}\right)\right|_{t=-1}\tag9\\\\ &=4\sqrt\pi\left(-\frac{\sqrt\pi (-\gamma-\log(4))}{1}+\frac{(1)\sqrt\pi (-\gamma)}{(1)^2}\right)\tag{10}\\\\ &=4\pi \log(4) \end{align}$$
And we are done!
NOTES:
In arriving at the right-hand side of $(1)$, we enforced the substitution $x\mapsto \log(x)$
In going from $(1)$ to $(2)$, we enforced the substitution $x\mapsto x+1$
In going from $(2)$ to $(3)$, we integrated by parts with $u=\log^2(x+1)$ and $v=-2x^{-1/2}$
In going from $(3)$ to $(4)$, we employed Feynman's trick.
In going from $(4)$ to $(5)$, we enforced the substitution $x\mapsto x^2$
In going from $(5)$ to $(6)$, we enforced the substitution $x\mapsto \tan(x)$
In going from $(6)$ to $(7)$, we made use of an integral representation of the Beta function
In going from $(7)$ to $(8)$, we made use of the relationship betwen the Beta function and the Gamma function
In going from $(8)$ to $(9)$, we took the derivative and used the relationship $\Gamma'(z)=\Gamma(z)\psi(z)$, we $\psi(z)$ is the Digamma function.
In arriving at $(10)$ we evaluated the derivative at $t=-1$ and used the fact that $\Gamma(1/2)=\sqrt\pi$, $\Gamma(1)=1$, $\psi(1/2)=-\gamma-\log(4)$, and $\psi(1)=-\gamma$
BONUS:
I thought that I would present the companion solution, which is solution to the currently posted question. Using the same approach as presented in the previous solution we have
$$\begin{align} \int_0^\infty \frac{xe^{-2x}}{(e^x-1)^{1/2}}\,dx&=\int_1^\infty \frac{\log(x)}{x^3(x-1)^{1/2}}\,dx\\\\ &=\int_0^\infty \frac{\log(x+1)}{(x+1)^3x^{1/2}}\,dx\\\\ &=\frac{d}{dt}\left.\left(\int_0^\infty \frac{(x+1)^t}{x^{1/2}}\,dx\right)\right|_{t=-3}\\\\ &=\frac{d}{dt}\left.\left(2\int_0^\infty (x^2+1)^t\,dx\right)\right|_{t=-3}\\\\ &=\frac{d}{dt}\left.\left(2\int_0^{\pi/2} \sec^{2t+2}(x)\,dx\right)\right|_{t=-3}\\\\ &=\frac{d}{dt}\left.\left(B(1/2,-t-1/2)\right)\right|_{t=-3}\\\\ &=\frac{d}{dt}\left.\left(\frac{\Gamma(1/2)\Gamma(-t-1/2)}{\Gamma(-t)}\right)\right|_{t=-3}\\\\ &=\sqrt\pi\left.\left(-\frac{\Gamma(-t-1/2)\psi(-t-1/2)}{\Gamma(-t)}\right)\right|_{t=-3}\\\\ &+\sqrt\pi\left.\left(\frac{\Gamma(-t)\Gamma(-t-1/2)\psi(-t)}{\Gamma^2(-t)}\right)\right|_{t=-3}\\\\ &=\sqrt\pi\left(-\frac{(3\sqrt\pi/4) (8/3-\gamma-\log(4))}{2}+\frac{(2)(3\sqrt\pi/4) (3/2-\gamma)}{(2)^2}\right)\\\\ &=\frac{\pi}{16}\left( 6\log(4)-7\right) \end{align}$$
And we are done (again)!