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.