How to find $\int_{0}^{\infty}\frac{x^{2}}{e^{x}+1}dx$ using known result for $\int_{0}^{\infty}\frac{x^{2}}{e^{x}-1}dx$ I need to find integral of $\int_{0}^{\infty}\frac{x^{2}}{e^{x}+1}dx$. From
this Riemann_zeta_function , it gives $\zeta\left(  s\right)  =\frac{1}{\Gamma\left(
s\right)  }\int_{0}^{\infty}\frac{x^{s-1}}{e^{x}-1}dx$. Hence for $s=3$ this gives
\begin{align}
\int_{0}^{\infty}\frac{x^{2}}{e^{x}-1}dx  & =\zeta\left(  3\right)
\Gamma\left(  3\right)  \nonumber\\
& =2\zeta\left(  3\right)  \tag{1}%
\end{align}
My question is, how to use the above result to find integral of $\int%
_{0}^{\infty}\frac{x^{2}}{e^{x}+1}dx$? Mathematica gives
\begin{equation}
\int_{0}^{\infty}\frac{x^{2}}{e^{x}+1}dx=\frac{3}{2}\zeta\left(  3\right)
\tag{2}
\end{equation}
I tried to see if there is change of variable or other trick, but I do not
see one. I do know about a direct method given in this answer here, but I want
to see if it is possible, with some trick, to use the result from (1) to find (2).

Let $\displaystyle \mathcal I =\int_{0}^{\infty}\frac{x^{2}}{e^{x}-1}\,\mathrm{d}x $ and $\displaystyle \mathcal J = \int_{0}^{\infty}\frac{x^{2}}{e^{x}+1}\,\mathrm{d}x. $
You already know $\mathcal I =2\zeta\left(  3\right)$ but you seek $\mathcal J$.
$\displaystyle \mathcal I =\int_{0}^{\infty}\frac{(2x)^{2}}{e^{2x}-1}\,\mathrm{d}(2x) = \int_0^\infty \frac{8x^2}{e^{2x}-1}\, \mathrm{d}x$
But  $\displaystyle \frac{8x^2}{e^{2x}-1} = \frac{4x^2}{e^x-1}-\frac{4x^2}{e^x+1}$
Thus $\displaystyle \mathcal I =  4\int_{0}^{\infty} \frac{x^2}{e^x-1}\,\mathrm{d}x-4\int_{0}^{\infty} \frac{x^2}{e^x+1}\,\mathrm{d}x $
So  $\mathcal I = 4 \mathcal I-4 \mathcal J $ and it follows that $\displaystyle \mathcal J =\frac{3}{2}\zeta(3). $
