Compute variance of logistic distribution

Hint. You may write $$\begin{align} \int_{-\infty}^{\infty}\frac{x^2e^{-x}}{(1+e^{-x})^2}dx &=\int_{-\infty}^{0}\frac{x^2e^{-x}}{(1+e^{-x})^2}dx +\int_{0}^{\infty}\frac{x^2e^{-x}}{(1+e^{-x})^2}dx\\\\ &=\int_{0}^{\infty}\frac{x^2e^{x}}{(1+e^{x})^2}dx +\int_{0}^{\infty}\frac{x^2e^{-x}}{(1+e^{-x})^2}dx\\\\ &=\int_{0}^{\infty}\frac{x^2e^{x}}{e^{2x}(1+e^{-x})^2}dx +\int_{0}^{\infty}\frac{x^2e^{-x}}{(1+e^{-x})^2}dx\\\\ &=2\int_{0}^{\infty}\frac{x^2e^{-x}}{(1+e^{-x})^2}dx\\\\ &=2\int_{0}^{\infty}x^2\sum_{n=1}^\infty n(-1)^{n-1}e^{-nx} dx\\\\ &=2\sum_{n=1}^\infty n(-1)^{n-1}\int_{0}^{\infty}x^2e^{-nx} dx\\\\ &=2\sum_{n=1}^\infty n(-1)^{n-1}\frac2{n^3}\\\\ &=4\sum_{n=1}^\infty (-1)^{n-1}\frac1{n^2}\\\\ &=\frac{\pi^2}3 \end{align} $$ where the interchange between sum and integral is easy to justify and where we have used some standard evaluations.

I might as well add my answer as well

$$V=\int_{-\infty}^{\infty}\frac{x^2e^{-x}}{(1+e^{-x})^2}dx= \int_{0}^{1}\Bigg(\ln\bigg(\frac{p}{1-p}\bigg)\Bigg)^2dp=\frac{\pi^2}{3}$$

To prove the last identity split the logarithm into three parts

$$ \Bigg(\ln\bigg(\frac{p}{1-p}\bigg)\Bigg)^2 = \log^2(p) - 2 \log p \log (1-p) + \log^2(1-p) $$ Because of symmetry the first and last integrals are equal, and are easilly calculated by parts

$$ \begin{align*} \int_0^1 \log^2p + \log^2(1-p)\,\mathrm{d}p & =2\int_0^1 \log^2p \,\mathrm{d}p \\ & =\bigg[ p \log^2p\bigg]_0^1 - 2 \int_0^1 p\frac{2 \log p}{p}\,\mathrm{d}p = -4 \int_0^1 \log p \,\mathrm{d}p = 4 \end{align*} $$ So the only remaining problem is the middle integral. However this canbe calculated using the series expansion of $\log(1-p)$ since $p \in (0,1)$. \begin{align*} \int_0^1 \log p \log(1-p)\,\mathrm{d}p & = \int_0^1 \log p \left( \sum_{n=1}^\infty \frac{p^n}{n} \right) \,\mathrm{d}p \\ & = \sum_{n=1}^\infty \frac{1}{n} \int_0^1 p^n \cdot \log p \,\mathrm{d}p \\ & = \sum_{n=1}^\infty \frac{1}{n} \left\{ \left[ \frac{p^{n+1}\log p}{p+1} \right]_0^1 - \int_0^1 \frac{p^n}{n+1} \,\mathrm{d}p \right\} \\ & = \sum_{n=1}^\infty \frac{1}{n} \cdot \frac{1}{(n+1)^2} \\ & = \left( \sum_{n=1}^\infty \frac{1}{n} - \frac{1}{n+1} \right) - \left( -\frac{1}{(0+1)^2} + \sum_{n=0}^\infty \frac{1}{(n+1)^2} \right) \\ & = 1 + 1 - \sum_{n=1}^\infty \frac{1}{n^2} = 2 - \frac{\pi^2}{6} \end{align*} Where the last sum is well know.