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).
 A: 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). $
A: Well using the fact that:
$$(a+1)(a-1)=a^2-1$$
we can say that:
$$\frac{1}{e^{2x}-1}=\frac12\frac{1}{e^x-1}-\frac12\frac{1}{e^x+1}$$
$$\therefore \frac{1}{e^x+1}=\frac{1}{e^x-1}-\frac{2}{e^{2x}-1}$$
so In term of your integral we can say:
$$I=\int_0^\infty\frac{x^2}{e^x+1}dx=\int_0^\infty\frac{x^2}{e^{x}-1}dx-2\int_0^\infty\frac{x^2}{e^{2x}-1}dx\tag{1}$$
$$I_1=\int_0^\infty\frac{x^2}{e^x-1}dx\tag{2}$$
Now, the first integral you know the value of so lets look at the second, using the substitution $u=2x\Rightarrow dx=\frac{du}2$ we can say the following:
$$\int_0^\infty\frac{x^2}{e^{2x}-1}dx=\int_0^\infty\frac{(u/2)^2}{e^u-1}\frac{du}2=\frac18\int_0^\infty\frac{u^2}{e^u-1}du=\frac18I_1\tag{3}$$

Now from this we can say:
$$I=I_1-\frac28I_1=\frac34I_1\tag{4}$$
which is the same as saying:
$$\int_0^\infty\frac{x^2}{e^x+1}dx=\frac34\int_0^\infty\frac{x^2}{e^x-1}dx\tag{5}$$
