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

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Answer 1 · 2017-02-08T12:26:45

$1/4ln|(x+2)/(x-2)|+C$ .

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

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

$=1/4int{(2+x)+(2-x)}/((2+x)(2-x))dx$

$=1/4int[cancel((2+x))/{cancel((2+x))(2-x)}+cancel((2-x))/{cancel((2-x))(2+x)}]$

$=1/4int(-1/(x-2)+1/(x+2))dx$

Knowing that, for $a!=0, int1/(ax+b)dx=1/aln|ax+b|+c,$

$I=1/4(-ln|x-2|+ln|x+2|)$

$=1/4ln|(x+2)/(x-2)|+C$ .

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