Evaluate the integral $\int_{-1}^1\frac{dx}{(e^x+1)(x^2+1)}$.
Evaluate:
$$\int_{-1}^1\frac{dx}{(e^x+1)(x^2+1)}$$
Showing that $\int\limits_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm dx = \int\limits_0^a f(x) \mathrm dx$, when $f$ is even
From the above, we get that $$\int_{-1}^{1} \dfrac{dx}{(1+e^x)(1+x^2)} = \int_{0}^{1} \dfrac{dx}{(1+x^2)} = \arctan(x) \left. \right\vert_{x=0}^{1} = \dfrac{\pi}4$$