What is $\int\frac{x^4}{1+ e^x} dx$? Here's the integral I have,

$$ \displaystyle\int\dfrac{x^4}{1+ e^x} dx $$

I tried the usual methods I know, but I failed miserably.
How would you all approach this problem?

$\int\frac {x^4}{1+e^x} \ dx\\
\int\frac {x^4e^{-x}}{1+e^{-x}} \ dx$
Converting to a geometric series: $\frac {y}{1+y} = \sum_\limits{n=1}^\infty (-y)^n$
$\int x^4\sum_\limits{n=1}^\infty (-1)^ne^{-nx} \ dx$
considering just one term $\int x^4e^{-nx} = (\frac {x^4}{n} + \frac {4x^3}{n^2} + \frac {12x^2}{n^2} + \frac {24x}{n^3} + \frac {24}{n^4}) e^{-nx}$
$\sum_\limits{n=1}^{\infty}(-1)^n(\frac {x^4}{n} + \frac {4x^3}{n^2} + \frac {12x^2}{n^2} + \frac {24x}{n^3} + \frac {24}{n^4}) e^{-nx}$
It is worth noting
$\frac{1}{\Gamma(s)} \int_0^{\infty} \frac {x^{s-1}}{e^x -1} \ dx = \zeta(s)$
