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Question
Integrate the following with respect to x.
`1/(9 - 16x^2)`
Solution
`int 1/(9 - 16x^2) "d"x`
= `int 1/(16[9/16 - x^2]) "d"x`
= `1/16 int ("d"x)/((3/4)^2 - (x)^2`
= `1/16[1/(2^(3/4)) log |(3/4 + x)/(3/4 - x)|] + "c"`
= `1/16 [1/(3/2) log |(3 + 4x)/(3 - 4x)|] + "c"`
= `1/24 log |(3 + 4x)/(3 - 4x)| + "c"`
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