How do you integrate #int (1-x^2)/((x+1)(x-6)(x-3)) # using partial fractions?

2 Answers
Jul 17, 2018

The answer is #=-5/3ln(|x-6|)+2/3ln(|x-3|)+C#

Explanation:

First, simplify

#(1-x^2)/((x+1)(x-6)(x-3))=((1-x)cancel(1+x))/(cancel(x+1)(x-6)(x-3))#

Therefore,

#(1-x)/((x-6)(x-3))=A/(x-6)+B/(x-3)#

#=(A(x-3)+B(x-6))/((x-6)(x-3))#

The denominators are the same, compare the numerators

#1-x=A(x-3)+B(x-6)#

Let #x=6#, #-5=3A#, #=>#, #A=-5/3#

Let #x=3#, #-2=-3B#, #=>#, #B=2/3#

Therefore,

#(1-x)/((x-6)(x-3))=(-5/3)/(x-6)+(2/3)/(x-3)#

Therefore, the integral is

#I=int((1-x)dx)/((x-6)(x-3))=int(-5/3dx)/(x-6)+int(2/3dx)/(x-3)#

#=-5/3ln(|x-6|)+2/3ln(|x-3|)+C#

Jul 17, 2018

# 2/3ln|(x-3)|-5/3ln|(x-6)|+C, #

# or, ln|(x-3)^(2/3)/(x-6)^(5/3)|+C.#

Explanation:

We use the method of partial fraction in disguise.

Let, #I=int(1-x^2)/{(x+1)(x-6)(x-3)}dx#,

#=int{(1+x)(1-x)}/{(x+1)(x-6)(x-3)}dx#.

#:. I=int(1-x)/{(x-6)(x-3)}dx#,

Note that,

#(x-3)-(x-6)=3, and, 2(x-3)-(x-6)=x#.

#:. 1-x=1/3(3)-x#,

#=1/3{(x-3)-(x-6)}-{2(x-3)-(x-6)}#,

#=(1/3-2)(x-3)+(1-1/3)(x-6)#,

#=2/3(x-6)-5/3(x-3)#.

#:. I=int{2/3(x-6)-5/3(x-3)}/{(x-6)(x-3)}dx#,

#=2/3int(x-6)/{(x-6)(x-3)}dx-5/3int(x-3)/{(x-6)(x-3)}dx#,

#=2/3int1/(x-3)dx-5/3int1/(x-6)dx#,

#rArr I=2/3ln|(x-3)|-5/3ln|(x-6)|+C, #

# or, I=ln|(x-3)^(2/3)/(x-6)^(5/3)|+C.#