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#