Question

How do you integrate `(3x+11)/(x^2-x-6)` using partial fractions?

Answer 1

The answer is `4ln|x- 3| - ln|x+ 2| + C`

Explanation:

The denominator can be factored as `(x - 3)(x + 2)`, so:

`A/(x- 3) + B/(x + 2) = (3x + 11)/((x- 3)(x + 2))`

`A(x + 2) + B(x- 3) = 3x +11`

`Ax + 2A + Bx - 3B = 3x +11`

`(A + B)x + (2A - 3B) = 3x + 11`

We now write a systems of equations to solve for `A` and `B`.

`{(A + B = 3), (2A - 3B = 11):}`

Solving:

`B = 3 - A -> 2A - 3(3 - A) = 11 -> 2A - 9 + 3A = 11 ->5A = 20-> A= 4`

`A + B = 3 -> 4 + B = 3 -> B = -1`

So, `A = 4` and `B = -1`.

The partial fraction decomposition is therefore `color(blue)(4/(x- 3) - 1/(x + 2))`.

This can be integrated using the rule `int1/xdx = ln|x| + C`.

`int(4/(x- 3) - 1/(x + 2)) = 4ln|x- 3| - ln|x+ 2| + C`

Hopefully this helps!