Thread: Simple Integral

1. Simple Integral

So i'm having trouble with a integral that i've been seeing alot lately..It goes like $\displaystyle \frac{1}{A+x^2}$ where A could be anything that is not 1...Now the solution ends up being arctan. However i've been struggling to figure out how to substitute in U to get the right answer...for example $\displaystyle \frac{1}{8+x^2}$ . What should i substitute?

2. In the integral

$\displaystyle \int\frac{dx}{{a^2+x^2}}$

By Trigonometric Substitution:

$\displaystyle x=a\tan(\theta),\quad dx=a\sec^2(\theta)\,d\theta$

$\displaystyle \theta=\arctan\left(\frac{x}{a}\right)$
so that the integral becomes

$\displaystyle \qquad \int\frac{dx}{{a^2+x^2}}$
= [Math]\int\frac{a\sec^2(\theta)\,d\theta}{{a^2+a^2\tan^2 (\theta)}}[/tex]
= $\displaystyle \int\frac{a\sec^2(\theta)\,d\theta}{{a^2(1+\tan^2( \theta))}}$
= $\displaystyle \int \frac{a\sec^2(\theta)\,d\theta}{{a^2\sec^2(\theta) }}$
$\displaystyle = \int \frac{d\theta}{a} = \frac{\theta}{a}+C$
$\displaystyle = \frac{1}{a} \arctan \left(\frac{x}{a}\right)+C$

(provided ''a'' > 0).

The long and short of it is when you have an integral of the form $\displaystyle \int\frac{dx}{{a^2+x^2}}$ you get an result of the form: $\displaystyle \frac{1}{a} \arctan \left(\frac{x}{a}\right)+C$

It's a good one to memorize.