How to find $ \int \frac{x^8}{e^{2 x}+1} \, dx$?

Question

I need help with this $$ \int \frac{x^8}{e^{2 x}+1} \, dx$$ Can't figure it out. I tried with partial integration, and I lose myself.

If someone can help me. Thanks a lot.

Answer 1

Case $1$: $x\leq0$

Then $\int\dfrac{x^8}{e^{2x}+1}~dx$

$=\int\sum\limits_{n=0}^\infty(-1)^nx^8e^{2nx}~dx$

$=\int\left(x^8+\sum\limits_{n=1}^\infty(-1)^nx^8e^{2nx}\right)dx$

$=\dfrac{x^9}{9}+\sum\limits_{n=1}^\infty\sum\limits_{k=0}^8\dfrac{(-1)^{n-k}8!2^{k-9}n^{k-9}x^ke^{2nx}}{k!}+C$

Case $2$: $x\geq0$

Then $\int\dfrac{x^8}{e^{2x}+1}~dx$

$=\int\dfrac{x^8e^{-2x}}{e^{-2x}+1}~dx$

$=\int\sum\limits_{n=0}^\infty(-1)^nx^8e^{-2(n+1)x}~dx$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^8\dfrac{(-1)^{n+1}8!2^{k-9}(n+1)^{k-9}x^ke^{-2(n+1)x}}{k!}+C$

Answer 2

Can't figure it out.

No wonder you can't, since it's not expressible in terms of elementary functions. We do, however, have the following two identities: $$\int_0^\infty\frac{x^n}{e^x-1}~dx~=~n!~\zeta(n+1)$$ and $$\int_0^\infty\frac{x^n}{e^x+1}~dx~=~n!~\eta(n+1),$$ see the Riemann $\zeta$ and Dirichlet $\eta$ functions for more details.