<s>[INST] Simple generalized integral The integral to compute is $\displaystyle\int_0^\infty \frac{1}{3+x^2} \, \mathrm dx$.
I know how to compute the indefinite integral of this function its gives me
$$\frac{\sqrt{3}}{3} \arctan\left(\frac{x}{\sqrt{3}}\right).$$
When i compute the definite integral it now gives me :
$$\frac{\sqrt{3}}{3} \left(\lim\limits_{x \to \infty }\arctan\left(\frac{x}{\sqrt{3}}\right)-\arctan(0)\right).$$
Then i don't understand why my teacher writes that arctan(0)=0 because it also can be equal to π, and more stranger he found i don't know how that $\lim\limits_{x \to \infty }\arctan\left(\frac{x}{\sqrt{3}}\right)=\pi/2$. Thank you for help !
EDIT : I only need how to compute the limit now.
 [/INST] The way i would approach this problem is . 
$$ \int_0^\infty \frac{1}{3+x^2}dx $$
I would divide the $$ x^2+3 $$ by 3 to factor out the constants. 
$$ \int_0^\infty \frac{1}{3(\frac{x^2}{3})+1}dx $$
then we can substitute our function with
$ u^2=\frac{x^2}{3}$
$u= \frac{x}{\sqrt{3}} $
The integral of
$$ \int_0^\infty \frac{1}{3(\frac{x^2}{3})+1}dx $$
$du= \frac{1}{\sqrt{3}}dx$
When we substitute we get 
$$ \frac{1}{\sqrt{3}}\int_0^\infty \frac{1}{{u^2}+1}dx $$
the integral would result to 
$$ \frac{1}{\sqrt{3}}\int_0^\infty \frac{1}{{u^2}+1}dx =\frac{1}{\sqrt{3}}arctan(u)|_0^\infty $$
Then we substitute back $u= \frac{x}{\sqrt{3}} $
This would give us the step before the final answer.
$$\frac{1}{\sqrt{3}}arctan(\frac{x}{\sqrt{3}})|_0^\infty $$
then we can evaluate the value and the limit 
$$\frac{1}{\sqrt{3}}\left(\lim\limits_{a \to \infty }\arctan\left(\frac{a}{\sqrt{3}}\right)-\arctan(0)\right)= (\frac{pi}{2\sqrt{3}}) $$
because as "a" approaches infinity arctan(a/sqrt(3))  = Pi/2 and arctan(0/sqrt(3)) = 0 
$$ \frac{1}{\sqrt{3}}( \frac{pi}{2} - 0 ) = (\frac{pi}{2\sqrt{3}}) $$
Thats the final correct answer and evaluation.
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