Question

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

Answer 1

`-1/6lnabs(x+3)-lnabs(x-2)+7/6lnabs(x-3)+C`

Explanation:

First, focus on writing the fraction in terms of partial fractions:

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

Multiply both sides by `(x+3)(x-2)(x-3)`:

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

Expand:

`2x+1=A(x^2-5x+6)+B(x^2-9)+C(x^2+x-6)`

Sort by degree:

`2x+1=x^2(A+B+C)+x(-5A+C)+(6A-9B-6C)`

Giving the system:

`{(A+B+C=0),(-5A+C=2),(6A-9B-6C=1):}`

Solving which gives:

`{(A=-1/6),(B=-1),(C=7/6):}`

Thus, we know that

`int(2x+1)/((x+3)(x-2)(x-3))dx=int-1/(6(x+3))-1/(x-2)+7/(6(x-3))dx`

Splitting up the integral and refactoring constants:

`=-1/6int1/(x+3)dx-int1/(x-2)dx+7/6int1/(x-3)dx`

`=-1/6lnabs(x+3)-lnabs(x-2)+7/6lnabs(x-3)+C`