How do you evaluate the integral $int dx/(4+x^2)$ ?

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Answer 1 · 2017-01-15T14:18:25

You may know the following rule:

$intdx/(a^2+x^2)=1/aarctan(x/a)+C$

So:

$intdx/(4+x^2)=intdx/(2^2+x^2)=1/2arctan(x/2)+C$

We can derive the $intdx/(a^2+x^2)$ rule. To do so, let $x=atantheta$ .

Thus, $dx=asec^2thetad theta$ and $x^2=a^2tan^2theta$ .

$intdx/(a^2+x^2)=int(asec^2thetad theta)/(a^2+a^2tan^2theta)$

Factoring the $a^2$ terms and using the identity $tan^2theta+1=sec^2theta$ :

$=int(asec^2thetad theta)/(a^2(1+tan^2theta))=int(sec^2thetad theta)/(asec^2theta)=1/aintd theta$

The antiderivative of $1$ is $theta$ , since we're working in terms of $theta$ here:

$=1/atheta+C$

From $x=atantheta$ , our original substitution, we can solve for $theta$ . Note that $tantheta=x/a$ so $theta=arctan(x/a)$ . Then:

$=1/aarctan(x/a)+C$

Which is the above stated rule.

How do you evaluate the integral int dx/(4+x^2)int dx/(4+x^2)?