Definite integration that results in inverse trigonometric functions

Question

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.

Answer 1

$$ \arctan\sqrt3 - \arctan 1 = \frac \pi 3 - \frac \pi 4 = \frac{4\pi}{12} - \frac{3\pi}{12} = \frac{(4-3)\pi}{12} = \frac \pi {12}. $$ One of a number of ways to see that this need not involve any of the "nonprincipal" values of the arctangent is this: $$ \text{If } 1 \le x \le \sqrt 3 \text{ then } \frac 1 2 \ge \frac 1 {1+x^2} \ge \frac 1 4, $$ $$ \text{so } \frac{\sqrt 3-1}2 \ge \int_1^{\sqrt 3} \frac{dx}{1+x^2} \ge \frac {\sqrt3-1} 4. $$

Answer 2

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 $\arctan_1(x) = \arctan(x) + k\pi$.