I wish to integrate $\int_{-a}^a \frac{dx}{1+e^x}.$ By symmetry, the above is equal to $\int_{-a}^a \frac{dx}{1+e^{-x}}$ Now multiply by $e^x/e^x$ to get $\int_{-a}^a \frac{e^x}{1+e^x} dx$ which integrates to $\log(1+e^x) |^a_{-a} = \log((1+e^a)/(1+e^{-a})),$ which is not correct. According to Wolfram, we should get $2a + \log((1+e^{-a})/(1+e^a)).$ Where is the mistake?
EDIT: Mistake found: was using log on calculator, which is base 10.