Integral from thermodynamics I've been trying to solve analytically an integral I stumbled into while working in a thermodynamics problem. I could solve it numerically, but it would be nicer if I can find an actual solution.
I've been trying to find something but I'm stuck, so if anyone knows a technique I could use, it would be greatly appreciated.
Enthalpy can be defined as: 
$$h = \int c_p(T) dT$$
But:
$$c_p(T) = c_{pp}\cdot \left\{ 1 + \frac{\gamma_p - 1}{\gamma_p}\left[ \left(\frac{\Theta}{T}\right)^2 \cdot \frac{e^{\Theta/T}}{(e^{\Theta/T}-1)^2} \right] \right\}$$
Where $c_{pp}, \gamma_p$ and $\Theta$ are constants. Thus:
$$h = c_{pp} \cdot\left[ T + \frac{\gamma_p-1}{\gamma_p} \int \left(\frac{\Theta}{T}\right)^2 \cdot \frac{e^{\Theta/T}}{(e^{\Theta/T} - 1)^2}dT \right]$$
The integral in question would then be:
$$ \int \left(\frac{\Theta}{T}\right)^2 \cdot \frac{e^{\Theta/T}}{(e^{\Theta/T} - 1)^2}dT$$
Here it is also in another form if it helps:
$$ \Theta ^2 \int \left( \frac{1}{T^2} \cdot \frac{1}{(e^{\Theta/T} - 1)} +  \frac{1}{T^2} \cdot \frac{1}{(e^{\Theta/T} - 1)^2} \right)dT$$
Thanks in advance!

Your integral is $\int\frac{\Theta^2}{4T^2}\operatorname{csch}^2\frac{\Theta}{2T}dT=\frac{\Theta}{2}\coth\frac{\Theta}{2T}+C$.
