Question

How do you integrate `int(x+1)/((x-9)(x+3)(x-2))` using partial fractions?

Answer 1

Set up a partial fraction decomposition and choose `x`-values to find the numerators of the partial fractions to get `int(x+1)/((x-9)(x+3)(x-2))dx=5/42lnabs(x-9)-1/30lnabs(x+3)-3/35lnabs(x-2)+C`.

Explanation:

Since all the factors in the denominator of `(x+1)/((x-9)(x+3)(x-2))` are linear (yay!), our partial fraction decomposition will be:
`A/(x-9)+B/(x+3)+C/(x-2)`

For now, let's ignore the integral and focus on decomposing this fraction:
`(x+1)/((x-9)(x+3)(x-2))=A/(x-9)+B/(x+3)+C/(x-2)`

`x+1=A(x+3)(x-2)+B(x-9)(x-2)+C(x-9)(x+3)`

Let `x=9` to find the value of `A`:
`9+1=A(9+3)(9-2)+B(9-9)(9-2)+C(9-9)(9+3)`
`10=84A`
`A=10/84=5/42`

Let `x=-3` to find the value of `B`:
`-3+1=A(-3+3)(-3-2)+B(-3-9)(-3-2)+C(-3-9)(-3+3)`
`-2=60B`
`B=-2/60=-1/30`

And let `x=2` to find the value of `C`:
`2+1=A(2+3)(2-2)+B(2-9)(2-2)+C(2-9)(2+3)`
`3=-35C`
`C=-3/35`

Therefore, `(x+1)/((x-9)(x+3)(x-2))=(5/42)/(x-9)-(1/30)/(x+3)-(3/35)/(x-2)`. Our integral is now:
`int(5/42)/(x-9)-(1/30)/(x+3)-(3/35)/(x-2)dx`

Using the sum rule, this becomes:
`int(5/42)/(x-9)dx-int(1/30)/(x+3)dx-int(3/35)/(x-2)dx`

Integrating these gives:
`5/42int1/(x-9)dx-1/30int1/(x+3)dx-3/35int1/(x-2)dx`
`5/42lnabs(x-9)-1/30lnabs(x+3)-3/35lnabs(x-2)+C`