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.

Your solution is absolutely correct. $\log(\frac{1+e^a}{1+e^{-a}})=2a+\log(\frac{1+e^{-a}}{1+e^a})$

The authors might have arrived at the solution like this. Let $I_1=\int^{a}_{-a} \frac{1}{1+e^x}dx$  and $I_2=\int^{a}_{-a} \frac{e^x}{1+e^x}dx$
Adding $I_1$ and $I_2$ , $I_1+I_2=\int^{a}_{-a}dx=2a$ As you have calculated $I_2=\log\left(\frac{1+e^a}{1+e^{-a}}\right)$
Therefore $I_1=2a-I_2=2a-\log\left(\frac{1+e^a}{1+e^{-a}}\right)=2a+\log\left(\frac{1+e^{-a}}{1+e^{a}}\right) $

