# How do you find the antiderivative of int 1/(4-x^2)dx?

Feb 8, 2017

$\frac{1}{4} \ln | \frac{x + 2}{x - 2} | + C$.

#### Explanation:

I=int1/(4-x^2)dx=int1/{(2+x)(2-x)dx

=1/4int4/{(2+x)(2-x)dx

$= \frac{1}{4} \int \frac{\left(2 + x\right) + \left(2 - x\right)}{\left(2 + x\right) \left(2 - x\right)} \mathrm{dx}$

$= \frac{1}{4} \int \left[\frac{\cancel{\left(2 + x\right)}}{\cancel{\left(2 + x\right)} \left(2 - x\right)} + \frac{\cancel{\left(2 - x\right)}}{\cancel{\left(2 - x\right)} \left(2 + x\right)}\right]$

$= \frac{1}{4} \int \left(- \frac{1}{x - 2} + \frac{1}{x + 2}\right) \mathrm{dx}$

Knowing that, for $a \ne 0 , \int \frac{1}{a x + b} \mathrm{dx} = \frac{1}{a} \ln | a x + b | + c ,$

$I = \frac{1}{4} \left(- \ln | x - 2 | + \ln | x + 2 |\right)$

$= \frac{1}{4} \ln | \frac{x + 2}{x - 2} | + C$.