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Question
Integrate the following with respect to x.
`1/(2x^2 + 6x - 8)`
Solution
`int 1/(2x^2 + 6x - 8) = 1/2 int ("d"x)/(x^2 + 3x - 4)`
= `1/2 int ("d"x)/((x + 3/2)^2 - 9/4 - 4)`
= `1/2 int ("d"x)/((x + 3/2)^2 - 25/4)`
= `1/2 int ("d"x)/((x + 3/2)^2 - (5/2)^2)`
= `1/2 1/(2(5/2)) log|(x + 3/2 - 5/2)/(x + 3/2 + 5/2)| + "c"`
= `1/10 log |(x - 1)/(x + 4)| + "c"`
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