Evaluate:
$\int_{-1}^1\frac{dx}{(e^x+1)(x^2+1)}$
Evaluate:
$\int_{-1}^1\frac{dx}{(e^x+1)(x^2+1)}$
From the above, we get that $\int_{-1}^{1} \dfrac{dx}{(1+e^x)(1+x^2)} = \int_{0}^{1} \dfrac{dx}{(1+x^2)} = \arctan(x) \left. \right\vert_{x=0}^{1} = \dfrac{\pi}4$