Question

How do you integrate `(2x-3)/(x^2-5x+6)` using partial fractions?

Answer 1

Please see the explanation.

Explanation:

Do partial fraction decomposition:

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

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

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

Solve for A by letting `x = 2`:

`(2(2) - 3) = A(2 - 3)`

`-1 = A`

Solve for B by letting `x = 3`:

`(2(3) - 3) = B(3 - 2)`

3 = B

Check:

`-1/(x - 2) + 3/(x - 3) =`

`-1/(x - 2)(x - 3)/(x - 3) + 3/(x - 3)(x - 2)/(x - 2) =`

`(-x + 3)/((x - 2)(x - 3)) + (3x - 6)/((x - 3)(x - 2)) =`

`(2x - 3)/((x - 3)(x - 2))`

This checks.

`int (2x - 3)/(x^2 - 5x + 6)dx = 3int 1/(x - 3)dx - int1/(x - 2)dx`

`int (2x - 3)/(x^2 - 5x + 6)dx = 3ln|x - 3| - ln|x - 2| + C`