I want to integrate this function:
$$\int\dfrac{x^2}{e^x-1}dx$$
I used integration by parts formula to integrate it.
However I have reached somewhere where I got something like this:
$$\int\dfrac{e^xx^3}{(e^x-1)^2}dx$$
Now I cannot go further. Is there any way I can solve this problem from here? Thank you!!
On
$\int\dfrac{x^2}{e^x-1}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{B_nx^{n+1}}{n!}dx$ (with the formula in http://en.wikipedia.org/wiki/Bernoulli_number#Generating_function)
$=\sum\limits_{n=0}^\infty\dfrac{B_nx^{n+2}}{n!(n+2)}+C$
This has no elementary antiderivative. This involves polylogarithms. See here. As TylerHG also pointed out, you may want to check this.
If so inclined, then you could write the denominator $e^x - 1 = -1 + \sum_{n=0}^{\infty} \frac{x^n}{n!} = \sum_{n=1}^{\infty} \frac{x^n}{n!}$