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Question
Find the integrals of the following:
`1/(9x^2 - 4)`
Solution
`int ("d"x)/(9x^2 - 4) = 1/9 int ("d"x)/(x^2 - 4/9)`
= `1/9 int ("d"x)/(x^2 - (2/3)^2`
`int ("d"x)/(x^2 - "a"^2) = 1/(2"a") log |(x - "a")/(x + "a")| + "c"`
= `1/9 xx 1/(2 xx (2/3)) log |(x - 2/3)/(x + 2/3)| + "c"`
= `1/9 xx 1/(4/3) log |((3x - 2)/3)/((3x + 2)/3)| + "c"`
= `1/9 xx 3/4 log |(3x - 2)/(3x + 2)| + "c"`
= `1/12 log |(3x - 2)/(3x + 2)| + "c"`
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