How to solve $\int\limits_0^x {x^n \over{1 + e^x}} dx$?

The "source" of $\zeta$ is $\int_0^\color{red}{\infty}t^n(1+e^t)^{-1}\,dt=n!\eta(n\color{red}{+1})=n!(1-2^{-n})\zeta(n+1)$. To prove, use $$\frac{1}{1+e^t}=\frac{e^{-t}}{1+e^{-t}}=\sum_{k=1}^\infty(-1)^{k-1}e^{-kt}$$ (then integrate termwise, etc.). For $x>0$, your original integral is treated the same way: $$\int_0^x\frac{t^n\,dt}{1+e^t}=\int_0^\infty\frac{t^n\,dt}{1+e^t}-\int_x^\infty\frac{t^n\,dt}{1+e^t}.$$

The first integral on the RHS is just considered, and the second is equal to $$\int_0^\infty\frac{(x+t)^n}{1+e^{x+t}}\,dt=\sum_{k=0}^n\binom{n}{k}x^{n-k}\int_0^\infty\frac{t^k\,dt}{1+e^{x+t}}.$$

The last integral is handled like the way above, and is equal to $$\sum_{j=1}^\infty(-1)^{j-1}e^{-jx}\int_0^\infty t^k e^{-jt}\,dt=-k!\sum_{j=1}^\infty\frac{(-e^{-x})^j}{j^{k+1}}=-k!\operatorname{Li}_{k+1}(-e^{-x}).$$

Finally, $$\int_0^x\frac{t^n\,dt}{1+e^t}=n!\left((1-2^{-n})\zeta(n+1)+\sum_{k=0}^{n}\frac{x^{n-k}}{(n-k)!}\operatorname{Li}_{k+1}(-e^{-x})\right).$$