<s>[INST] 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.
 [/INST] Substitution $x \rightarrow -x$ gives $$\int_{-1}^{1}  \frac{dx}{(e^x+1)(x^2+1)}=\int_{-1}^{1}  \frac{dx}{(e^{-x}+1)(x^2+1)}=\int_{-1}^{1}  \frac{e^xdx}{(e^x+1)(x^2+1)}$$
Therefore
$$
\int_{-1}^{1}  \frac{dx}{(x^2+1)}=\int_{-1}^{1}  \frac{dx}{(e^x+1)(x^2+1)} + \int_{-1}^{1}  \frac{e^xdx}{(e^x+1)(x^2+1)}=2\int_{-1}^{1}  \frac{dx}{(e^x+1)(x^2+1)}
$$
Should be easy now.
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