Question

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

Answer 1

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`

Answer 2

`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.`