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.

Both answers are equal.  Split your answer into $\log(1+e^a)-\log(1+e^{-a})$, and write this as
$$\begin{align*}\log(e^a(1+e^{-a}))-\log(e^{-a}(1+e^a))&=\log e^a+\log(1+e^{-a})-\log e^{-a}-\log(1+e^a)\\
&=2a+\log((1+e^{-a})/(1+e^a))\end{align*}$$
