How do you integrate $int (5x - 10)/ (x^2+2x-35) dx$ using partial fractions?

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Answer 1 · 2018-03-22T06:55:24

The answer is $=15/4ln(|x+7|)+5/4ln(|x-5|)+C$

Perform the decomposition into partial fractions

$(5x-10)/(x^2+2x-35)=(5x-10)/((x+7)(x-5))=A/(x+7)+B/(x-5)$

$=(A(x-5)+B(x+7))/((x+7)(x-5))$

The denominators are the same, compare the numerators

$5x-10=A(x-5)+B(x+7)$

Let $x=-7$ , $=>$ , $-45=-12A$ , $=>$ , $A=45/12=15/4$

Let $x=5$ , $=>$ , $15=12B$ , $=>$ , $A=15/12=5/4$

Therefore,

$(5x-10)/(x^2+2x-35)=(15/4)/(x+7)+(5/4)/(x-5)$

And finally,

$int((5x-10)dx)/(x^2+2x-35)=int(15/4dx)/(x+7)+int(5/4dx)/(x-5)$

$=15/4ln(|x+7|)+5/4ln(|x-5|)+C$

How do you integrate int (5x - 10)/ (x^2+2x-35) dxint (5x - 10)/ (x^2+2x-35) dx using partial fractions?