Integrating $\int \frac{1}{1+e^x} dx$ 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.

Another way to solve this is to notice that
$$f(x) = \frac{1}{1+e^{-x}}$$
is the logistic sigmoid, which is the conversion from log-odds to probability.  So,
$$f(x) + f(-x) = 1$$
Then,
$$
\int_{-a}^a f(x)dx=\int_0^a (f(x) + f(-x))dx=a.
$$
