The value of ${\int }_{-\pi /2}^{\pi /2} \frac{x\sin x}{e^x+1}dx$ is equal to

$\begin{array}{c}I={\int }_{-\pi /2}^{\pi /2} \frac{x\sin x}{e^x+1}dx={\int }_{-\pi /2}^{\pi /2} \frac{(-x)\sin (-x)}{e^{-x}+1}dx\text{ (Prop. 11) }\\ ={\int }_{-\pi /2}^{\pi /2} \frac{(x\sin x)e^x}{e^x+1}dx\end{array}$

$\begin{array}{l} 2I={\int }_{-\pi /2}^{\pi /2} \frac{x\sin x}{e^x+1}+{\int }_{-\pi /2}^{\pi /2} \frac{e^x(x\sin x)}{e^x+1}dx\\ ={\int }_{-\pi /2}^{\pi /2} \frac{\left(e^x+1\right)(x\sin x)}{e^x+1}dx\\ ={\int }_{-\pi /2}^{\pi /2} x\sin xdx=2{\int }_0^{\pi /2} x\sin xdx\\ \left(\text{ Prop. }12\right)\\ =2\left[-{\left.x\cos x\right|}_0^{\pi /2}+{\int }_0^{\pi /2} \cos xdx\right]\\ ={\left.2\sin x\right|}_0^{\pi /2}=2\\ \Rightarrow I=1.\end{array}$