How do you integrate int (x+5)/((x+3)(x-2)(x-5)) ∫x+5(x+3)(x−2)(x−5) using partial fractions?

2 Answers
anton
Oct 19, 2016

(x+5)/((x+3)(x-2)(x-5))=A/(x+5)+B/(x-2)+C/(x-5)x+5(x+3)(x−2)(x−5)=Ax+5+Bx−2+Cx−5
x^2(A+B+C)-x(7A+2B+6C)+10A-15B-6C=x+5x2(A+B+C)−x(7A+2B+6C)+10A−15B−6C=x+5
A+B+C=0; 7A +2B+C=1; 10A-15B-6C=5A+B+C=0;7A+2B+C=1;10A−15B−6C=5
A=1/5; B=-1/5; C=0A=15;B=−15;C=0
int((x+5)/((x+3)(x-2)(x-5)))=int1/(5(x+3))-int1/(5(x-2))=1/5(ln(x+3)-ln(x-2))=1/5ln((x+3)/(x-2))∫(x+5(x+3)(x−2)(x−5))=∫15(x+3)−∫15(x−2)=15(ln(x+3)−ln(x−2))=15ln(x+3x−2)

Narad T.
Oct 19, 2016

=1/20ln(x+3)-7/15ln(x-2)+5/12ln(x-5)+C=120ln(x+3)−715ln(x−2)+512ln(x−5)+C

Explanation:

Let's do the partial fractions first
(x+5)/((x+3)(x-2)(x-5))=A/(x+3)+B/(x-2)+C/(x-5)x+5(x+3)(x−2)(x−5)=Ax+3+Bx−2+Cx−5
=(A(x-2)(x-5)+B(x+3)(x-5)+C(x+3)(x-2))/((x+3)(x-2)(x-5))=A(x−2)(x−5)+B(x+3)(x−5)+C(x+3)(x−2)(x+3)(x−2)(x−5)
so have the following
x+5=A(x-2)(x-5)+B(x+3)(x-5)+C(x+3)(x-2)x+5=A(x−2)(x−5)+B(x+3)(x−5)+C(x+3)(x−2)
let x=2x=2 so 7=0-15B+07=0−15B+0 so B=-7/15B=−715
let x=5x=5 so 10=0+0+24C10=0+0+24C => C=5/12C=512
let x=-3x=−3 so 2=40A+0+02=40A+0+0 => A=1/20A=120
finally int((x+5)dx)/((x+3)(x-2)(x-5))=1/20intdx/(x+3)-7/15intdx/(x-2)+5/12intdx/(x-5)∫(x+5)dx(x+3)(x−2)(x−5)=120∫dxx+3−715∫dxx−2+512∫dxx−5
=1/20ln(x+3)-7/15ln(x-2)+5/12ln(x-5)+C=120ln(x+3)−715ln(x−2)+512ln(x−5)+C

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