How do you find $int x/(x^2-x-2)dx$ using partial fractions?

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Answer 1 · 2016-10-02T22:41:33

$int x/(x^2-x-2) dx = 2/3 ln abs (x-2)+1/3 ln abs(x+1) + C$

$x/(x^2-x-2) = x/((x-2)(x+1)) = 2/(3(x-2))+1/(3(x+1))$

So:

$int x/(x^2-x-2) dx = int 2/(3(x-2))+1/(3(x+1)) dx$

$color(white)(int x/(x^2-x-2) dx) = 2/3 ln abs (x-2)+1/3 ln abs(x+1) + C$

How do you find int x/(x^2-x-2)dxint x/(x^2-x-2)dx using partial fractions?