How do you integrate int (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^2(A+B+C)-x(7A+2B+6C)+10A-15B-6C=x+5
A+B+C=0; 7A +2B+C=1; 10A-15B-6C=5
A=1/5; B=-1/5; 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))

Narad T.
Oct 19, 2016

=1/20ln(x+3)-7/15ln(x-2)+5/12ln(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)
=(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)
let x=2 so 7=0-15B+0 so B=-7/15
let x=5 so 10=0+0+24C => C=5/12
let x=-3 so 2=40A+0+0 => A=1/20
finally int((x+5)dx)/((x+3)(x-2)(x-5))=1/20intdx/(x+3)-7/15intdx/(x-2)+5/12intdx/(x-5)
=1/20ln(x+3)-7/15ln(x-2)+5/12ln(x-5)+C

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