How to prove $\int^{\pi}_{-\pi}[4\arctan(e^x)-\pi]\mathrm{d}x={0}$ How to prove $$\int^{\pi}_{-\pi}[4\arctan(e^x)-\pi]\mathrm{d}x={0}?$$ 
I'm trying to prove that the function is odd, but I just can't succeed.
 A: To show that the integrand is odd is the right idea. We can prove this by a simple geometrical argument as pointed out by Travis. 
Consider a right triangle with legs $1$ and $u$. The accute angles are $\arctan(u)$ and $\arctan(1/u)$ and these must sum up to $\frac{\pi}{2}$ so for $u>0$ we have the identity
$$\arctan(u) + \arctan(1/u) = \frac{\pi}{2} \implies \arctan(u) - \frac{\pi}{4} = -\left(\arctan(1/u) - \frac{\pi}{4}\right)$$
taking $u=e^x$ it follows that $\arctan(e^x) - \frac{\pi}{4}$ is an odd function.
A: A simple way is 
to notice that
$\arctan(z)+\arctan(\frac1{z})
=\arctan(\frac{z+\frac1{z}}{1-z\frac1{z}})
=\arctan(\frac{z+\frac1{z}}{0})
=\frac{\pi}{2}
$
so that
$\begin{array}\\
\int^{\pi}_{-\pi}(4\arctan(e^x)-\pi)dx
&=\int^{\pi}_{0}(4\arctan(e^x)-\pi)dx+\int^{0}_{-\pi}(4\arctan(e^x)-\pi)dx\\
&=\int^{\pi}_{0}(4\arctan(e^x)-\pi)dx-\int^{0}_{\pi}(4\arctan(e^{-x})-\pi)dx\\
&=\int^{\pi}_{0}(4\arctan(e^x)-\pi)dx+\int_{0}^{\pi}(4\arctan(e^{-x})-\pi)dx\\
&=\int^{\pi}_{0}(4\arctan(e^x)-\pi+4\arctan(e^{-x})-\pi)dx\\
&=\int^{\pi}_{0}(4(\arctan(e^x)+\arctan(e^{-x}))-2\pi)dx\\
&=\int^{\pi}_{0}(4(\frac{\pi}{2})-2\pi)dx\\
&=0\\
\end{array}
$
