Definite integration that results in inverse trigonometric functions I try to evaluate this integrat
$$\int_1^{\sqrt{3}}\frac{1}{1+x^2}dx$$
It seems simple.
$$\int_1^{\sqrt{3}}\frac{1}{1+x^2}dx=\arctan(\sqrt{3})-\arctan(1)$$
My question is what exact number it is?
Should it be $\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}$?
Or should it be $(k_1\pi+\frac{\pi}{3})-(k_2\pi+\frac{\pi}{4})=(k_1-k_2)\pi+\frac{\pi}{12}$?
$k_1,k_2=0,\pm 1,\pm2,... $
I think the definite integral should be A number, but there seems can be many numbers for the results.
I think I may miss some very basic concption here.
Thank you very much for help.
 A: $$
\arctan\sqrt3 - \arctan 1 = \frac \pi 3 - \frac \pi 4 = \frac{4\pi}{12} - \frac{3\pi}{12} = \frac{(4-3)\pi}{12} = \frac \pi {12}.
$$
One of a number of ways to see that this need not involve any of the "nonprincipal" values of the arctangent is this:
$$
\text{If } 1 \le x \le \sqrt 3 \text{ then } \frac 1 2 \ge \frac 1 {1+x^2} \ge \frac 1 4,
$$
$$
\text{so } \frac{\sqrt 3-1}2 \ge \int_1^{\sqrt 3} \frac{dx}{1+x^2} \ge \frac {\sqrt3-1} 4.
$$
A: We know that $\tan(x)$ is many one as its periodic with period $\pi$. As a result for defining an inverse we need to fix our range to a fixed length $\pi$.
We can define $\arctan_1(x) : \Bbb R \to (k\pi-\tfrac{\pi}{2}, k\pi + \tfrac{\pi}{2}), k \in \Bbb Z$. We get our standard $\arctan(x)$ function by putting $k=0$.
Now we see that the integral 
$$\begin{align}
\int_{1}^{\sqrt{3}} \dfrac{1}{1+x^2} dx &=\arctan_1(x) |_{1}^{\sqrt{3}}\\
 &= k\pi + \tfrac{\pi}{3} -(k\pi + \tfrac{\pi}{4}) \\
&= \frac{\pi}{12}
\end{align}$$
is independant of branch of definition. This is self evident since $\arctan_1(x) = \arctan(x) + k\pi$.
