How do you integrate (2x-82)/(x^2+2x-48) dx2x−82x2+2x−48dx using partial fractions?

1 Answer
Narad T.
Oct 15, 2016

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

Explanation:

int(2x-82)dx/(x^2+2x-48)∫(2x−82)dxx2+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−82x2+2x−48=2x−82(x−6)(x+8)=Ax−6+Bx+8=A(x+8)+B(x−6)(x−6)(x+8)
2x-82=A(x+8) +B(x-6)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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