How should I solve this definite Integral I need to solve the integral 
$\displaystyle \int _{-1}^{3} \left[\arctan \left(\dfrac{x}{x^2+1}\right) + \arctan \left(\dfrac{x^2 + 1}{x}\right)\right]dx$.
I  thought I could use the identity $\arctan(x) + \operatorname{arccot}(x) = \dfrac{\pi}{2}$ to simplify the integral, But that gives the answer as $2\pi$ while the actual answer is given as:
$\pi$; 
can anyone tell me how should I solve this definite integral?
Thank you.
 A: $\forall x\in\mathbb{R}^*,\,\arctan(x)+\arctan\left(\frac{1}{x}\right)=\mathrm{sgn}(x)\frac{\pi}{2}$, split the integral into two integrals :
$$ \int _{-1}^{3} \left[\arctan \left(\dfrac{x}{x^2+1}\right) + \arctan \left(\dfrac{x^2 + 1}{x}\right)\right]dx=\int _{-1}^{0} \left[\arctan \left(\dfrac{x}{x^2+1}\right) + \arctan \left(\dfrac{x^2 + 1}{x}\right)\right]dx+\int _{0}^{3} \left[\arctan \left(\dfrac{x}{x^2+1}\right) + \arctan \left(\dfrac{x^2 + 1}{x}\right)\right]dx $$
We then have $$\int _{0}^{3} \left[\arctan \left(\dfrac{x}{x^2+1}\right) + \arctan \left(\dfrac{x^2 + 1}{x}\right)\right]dx=\frac{3\pi}{2}$$ and $$\int _{-1}^{0} \left[\arctan \left(\dfrac{x}{x^2+1}\right) + \arctan \left(\dfrac{x^2 + 1}{x}\right)\right]dx=-\frac{\pi}{2}$$ so that $$ \int _{-1}^{3} \left[\arctan \left(\dfrac{x}{x^2+1}\right) + \arctan \left(\dfrac{x^2 + 1}{x}\right)\right]dx=\pi $$
A: Remember that
$$\arctan x+\arctan\frac1x=\begin{cases}\cfrac\pi2,&x>0\\{}\\-\cfrac\pi2,&x<0\end{cases}$$
so we can simply write your integral as
$$\int_{-1}^0-\frac\pi2\,dx+\int_0^3\frac\pi2\,dx=-\frac\pi2+\frac{3\pi}2=\pi$$
A: You can't use that identity here since it doesn't apply. The identity you should use here is $$\arctan x+\arctan y=\arctan\left(\frac{x+y}{1-xy}\right).$$
