How do you integrate $int (x-2x^2)/((x-6)(x-2)(x-5))$ using partial fractions?

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Answer 1 · 2016-07-19T22:10:20

Divide by $x-2$ first

$x/[(x-6)(x-5)] = x[1/(x-6) - 1/(x-5)]$
$= 6/(x-6) - 5/(x-5)$
And the integral I is
$6 log |x-6| - 5 log|x-5|$

How do you integrate int (x-2x^2)/((x-6)(x-2)(x-5)) int (x-2x^2)/((x-6)(x-2)(x-5)) using partial fractions?