How do you integrate $int (3x + 2) / [(x - 1)(x + 4)]$ using partial fractions?

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Answer 1 · 2016-02-21T19:24:02

We have that

$(3x+2)/[(x-1)(x+4)]=2/(x+4)+1/(x-1)$

Hence

$int (3x+2)/[(x-1)(x+4)]dx=int 2/(x+4)dx+int 1/(x-1)dx= 2*logabs(x+4)+logabs(x-1)+c$

How do you integrate int (3x + 2) / [(x - 1)(x + 4)] int (3x + 2) / [(x - 1)(x + 4)] using partial fractions?