In a lecture, we had the following integral:
$$C_{\text{V, mol}} = 9R\left(\frac{T}{T_D} \right)^3 \int_{0}^{x_D}\frac{x^4e^x}{\left( e^x-1\right)^2}dx,$$ where $x_D \equiv \frac{\hbar\omega_D}{k_BT}$, and $x = \frac{\hbar\omega}{k_BT}$ is the temperature. Now, we said that for $T\gg T_D \equiv \frac{\hbar\omega_D}{k_B}$, we could approximate $C_{\text{V, mol}}$ as follows: $$C_{\text{V, mol}} \approx 9R \left( \frac{T}{T_D} \right)^3 \int_{0}^{T_D/T}x^2dx.$$ However, I don't know how to arrive at the last approximation. Clarification would be appreciated. :-)
I think I got it more or less $$I=\int_{0}^{t}\frac{x^4e^x}{\left( e^x-1\right)^2}dx$$
Expand the integrand around $x=0$ $$\frac{x^4e^x}{\left( e^x-1\right)^2}=x^2-\frac{x^4}{12}+\frac{x^6}{240}-\frac{x^8}{6048}+\frac{x^{10}}{172800}+O\left(x^ {12}\right)$$
Edit
In terms of approximation of the integrand, much better is $\frac{12 x^2}{x^2+12}$.
The integral can be computed rigorously $$I=12 t^2 \text{Li}_2\left(e^t\right)-24 t \text{Li}_3\left(e^t\right)+24 \text{Li}_4\left(e^t\right)-\frac{e^t t^4}{e^t-1}+4 t^3 \log \left(1-e^t\right)-\frac{4 \pi ^4}{15}$$