Q:

$\arctan(x) + \arctan(1/x)$ integration

How do I integrate 
$$\int_{1/b}^{b} \frac {\arctan(x)+ \arctan\left(\frac{1}{x}\right)}{x} dx \text{ ?} $$

A:

Deriving $\arctan$, you'll see that $\arctan x+\arctan\frac1x$ is constant. Passing to the limit for $x\to\pm\infty$ it's easy to see that $\arctan x+\arctan\frac1x=\operatorname{sgn}(x)\frac{\pi}2$. Thus, supposing $b>0$, you'll have
$$
\int_\frac1b^b\frac{\arctan x+\arctan\frac1x}x\,dx=\frac{\pi}2\int_{\frac1b}^b\frac1x\,dx
$$
from this it should be easy, I guess.

