How do you integrate $int (3x-4)/((x-1)(x+3)(x-6))$ using partial fractions?

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How do you integrate $int (3x-4)/((x-1)(x+3)(x-6))$ using partial fractions?

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Answer 1 · 2016-04-01T03:58:05

Partial fraction decomposition is:
$A/(x-1)+B/(x+3)+C/(x-6)$
Cover up rule or simultaneously solving by equating coefficients yields $A=1/20$ , $B=-13/36$ and $C=14/45$

$int(3x-4)/((x-1)(x+3)(x-6))dx$
$=1/20int1/(x-1)dx-13/36int1/(x+3)dx+14/45int(1/x-6)dx$
$=1/20ln|x-1|-13/36ln|x+3|+14/45ln|x-6|+c$

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How do you integrate $int (3x-4)/((x-1)(x+3)(x-6))$ using partial fractions?

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How do you integrate int (3x-4)/((x-1)(x+3)(x-6)) int (3x-4)/((x-1)(x+3)(x-6)) using partial fractions?

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How do you integrate $int (3x-4)/((x-1)(x+3)(x-6))$ using partial fractions?