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

2 Answers
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))#

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#