Question

How do you integrate `int (1-x^2)/((x-9)(x+6)(x-2))` using partial fractions?

Answer 1

`int(1-x^2)/((x-9)(x+6)(x-2))dx`

`=-16/21lnabs(x-9)-7/24lnabs(x+6)+3/56lnabs(x-2)+C`

Explanation:

Conveniently, the denominator in this problem is already factored.

To apply partial fraction decomposition, we want to express

`int(1-x^2)/((x-9)(x+6)(x-2))dx`

in the form

`int(A/(x-9)+B/(x+6)+C/(x-2))dx`

So we can equate

`(1-x^2)/((x-9)(x+6)(x-2))=A/(x-9)+B/(x+6)+C/(x-2)`

Multiplying both sides by the denominator of the left, we get:

`1-x^2=A(x+6)(x-2)+B(x-9)(x-2)+C(x-9)(x+6)`

Now let's expand the parentheses and group similar terms:

`1-x^2=A(x^2+4x-12)+B(x^2-11x+18)+C(x^2-3x-54)`

`1-x^2=(A+B+C)x^2+(4A-11B-3C)x+(-12A+18B-54C)`

Now we can compare the coefficients on both sides, and write a system of three equations:

`A+B+C=-1`
`4A-11B-3C=0`
`-12A+18B-54C=1`

Solving this system (whichever way you choose) gives:

`A=-16/21`

`B=-7/24`

`C=3/56`

So now we can rewrite the integral as:

`int(-16/21(1)/(x-9)-7/24(1)/(x+6)+3/56(1)/(x-2))dx`

`rArr-16/21int1/(x-9)dx-7/24int1/(x+6)dx+3/56int1/(x-2)dx`

Integrating, we get:

`-16/21lnabs(x-9)-7/24lnabs(x+6)+3/56lnabs(x-2)+C`