$\int\limits_{-\pi / 4}^{\pi / 4} \frac{e^x(x \sin x)}{e^{2 x}-1} d x$ is equal to

0 2 e none of these

0

$I=\int\limits_{-\pi / 4}^{\pi / 4} \frac{e^x(x \sin x)}{e^{2 x}-1} d x$ ∵ property $\int\limits_{-a}^a f(x) d x=0$ (f (–x) = –f (x), odd function) Hence I = 0 Hence (1) is the correct answer.