Evaluating $\int_{-1}^{1} \frac{dx}{(e^x+1)(x^2+1)}$ I would like to evaluate the following integral:
$$
\int_{-1}^{1}  \frac{dx}{(e^x+1)(x^2+1)} 
$$
I tried various methods but without success.

Hint. One may write
$$
\begin{align}
\int_{-1}^{1}  \frac{dx}{(e^x+1)(x^2+1)} &=\int_{-1}^0  \frac{dx}{(e^x+1)(x^2+1)} +\int_0^{1}  \frac{dx}{(e^x+1)(x^2+1)} 
\\\\&=\int_0^1  \frac{dx}{(e^{-x}+1)(x^2+1)} +\int_0^{1}  \frac{dx}{(e^x+1)(x^2+1)} 
\\\\&=\int_0^1  \frac{e^x\:dx}{(e^{x}+1)(x^2+1)} +\int_0^{1}  \frac{dx}{(e^x+1)(x^2+1)} 
\\\\&=\int_0^1  \frac{(e^x+1)\:dx}{(e^{x}+1)(x^2+1)} 
\\\\&=\int_0^1  \frac{dx}{(x^2+1)} 
\end{align}
$$ then it is easier.
