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#.