How do you integrate #(2x-82)/(x^2+2x-48) dx# using partial fractions?

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How do you integrate #(2x-82)/(x^2+2x-48) dx# using partial fractions?

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Answer 1 · 2016-10-15T07:02:40

7ln(x+8)-5ln(x-6)+C

Explanation:

#int(2x-82)dx/(x^2+2x-48)#
#(2x-82)/(x^2+2x-48)=(2x-82)/((x-6)(x+8))=A/(x-6)+B/(x+8) =(A(x+8) +B(x-6))/((x-6)(x+8))#
#2x-82=A(x+8) +B(x-6)#
#if x=6 ; -70=14A ; A=-5#
#if x=-8 ; 98=-14B , B=7#
#int(2x-82)dx/(x^2+2x-48)=int7dx/(x+8)-int5dx/(x-6#
#=7ln(x+8)-5ln(x-6)+C#

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How do you integrate #(2x-82)/(x^2+2x-48) dx# using partial fractions?

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