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

A: 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,
$$
\int_{-a}^a f(x)dx=\int_0^a (f(x) + f(-x))dx=a.
$$
A: Another way to evaluate this integral is by substitution. In fact, let $u=e^x$. Then \begin{eqnarray*}
\int\frac{1}{1+e^x}dx&=&\int\frac{1}{u(u+1)}du=\int\left(\frac{1}{u}-\frac{1}{u+1}\right)du\\
&=&\ln u-\ln(u+1)+C=x-\ln(e^x+1)+C
\end{eqnarray*}
A: All of these answers are correct, but there is a more correct answer:
$$
\int_{-a}^a \frac{dx}{1+e^x} = a.
$$
(You should check that $\log((1+e^a)/(1+e^{-a})) = \log(e^a) = a$....) And you've basically found the proof already: set
$$
I_1 = \int_{-a}^a \frac{dx}{1+e^x} \text{ and } I_2 = \int_{-a}^a \frac{e^x\, dx}{1+e^x}.
$$
Then $I_1=I_2$ as you saw, but also
$$
I_1 + I_2 = \int_{-a}^a \bigg( \frac{1}{1+e^x} + \frac{e^x}{1+e^x} \bigg) \,dx = \int_{-a}^a 1\,dx = 2a.
$$
Therefore $I_1=I_2=a$. "anon" commented to this effect on the original question.
