Some Tricky Integral

If we define $A(a) \; =\; \int_{-\infty}^\infty \frac{e^{ax}}{(1 + e^x)^2}\,dx \; = \; \int_0^\infty \frac{u^{a-1}}{(1+u)^2}\,du \; = \; B(a,2-a) \; = \; \Gamma(a)\Gamma(2-a)$ for $a > 0$ , then $\begin{aligned} A'(a) & = \Gamma(a)\Gamma(2-a)\big\{\psi(a) - \psi(2-a)\big\} \\ A''(a) & = \Gamma(a)\Gamma(2-a)\big\{(\psi(a) - \psi(2-a))^2 + \psi'(a) + \psi'(2-a)\big\} \end{aligned}$ so that $\int_{-\infty}^\infty \frac{x^2 e^x}{(1 + e^x)^2}\,dx \; = \; A''(1) \; = \; 2\psi'(1) \; = \; \tfrac13\pi^2 \; = \; \boxed{3.2898681}$