分母の2乗をどうするかな？

\begin{align*} \int_{0}^{\infty}\frac{x^{2}}{\left(1+e^{x}\right)^{2}}dx & =\int_{0}^{\infty}\frac{x^{2}}{\left(1+e^{-x}\right)^{2}e^{2x}}dx\\ & =\int_{0}^{\infty}x^{2}e^{-x}\frac{d}{dx}\frac{1}{1+e^{-x}}dx\\ & =\int_{0}^{\infty}x^{2}e^{-x}\frac{d}{dx}\sum_{k=0}^{\infty}\left(-e^{-x}\right)^{k}dx\\ & =\int_{0}^{\infty}x^{2}e^{-x}\sum_{k=0}^{\infty}\left(-1\right)^{k}\left(-k\right)e^{-xk}dx\\ & =\sum_{k=0}^{\infty}\left(-1\right)^{k}\left(-k\right)\int_{0}^{\infty}x^{2}e^{-\left(k+1\right)x}dx\\ & =\sum_{k=0}^{\infty}\left(-1\right)^{k}\left(-k\right)\mathcal{L}_{x}\left[x^{2}H\left(x\right)\right]\left(k+1\right)\\ & =\sum_{k=0}^{\infty}\left(-1\right)^{k}\left(-k\right)\frac{2!}{\left(k+1\right)^{2+1}}\\ & =2\sum_{k=0}^{\infty}\left(-1\right)^{k+1}\frac{k}{\left(k+1\right)^{3}}\\ & =2\sum_{k=0}^{\infty}\left(-1\right)^{k+1}\frac{k+1-1}{\left(k+1\right)^{3}}\\ & =2\sum_{k=0}^{\infty}\left(-1\right)^{k+1}\left(\frac{1}{\left(k+1\right)^{2}}-\frac{1}{\left(k+1\right)^{3}}\right)\\ & =-2\sum_{k=1}^{\infty}\left(-1\right)^{k+1}\left(\frac{1}{k^{2}}-\frac{1}{k^{3}}\right)\\ & =-2\left(\eta\left(2\right)-\eta\left(3\right)\right)\\ & =-2\left(\left(1-\frac{1}{2^{1}}\right)\zeta\left(2\right)-\left(1-\frac{1}{2^{2}}\right)\eta\left(3\right)\right)\\ & =-\zeta\left(2\right)+\frac{3}{2}\zeta\left(3\right)\\ & =-\frac{\pi^{2}}{6}+\frac{3}{2}\zeta\left(3\right) \end{align*}