Question

How do you integrate `int (5x+12)/(x^2+5x+6)` using partial fractions?

Answer 1

`ln(x+3)+ln(x+2) +C`

Explanation:

To begin we must take:

`(5x+12)/(x^2+5x+6)` and decompose it into its partial fractions. First factorise the denominator and then split the fraction up as follows:

`(5x+12)/((x+3)(x+2)) = A/(x+3) +B/(x+2)`

Now, if we multiply the whole thing through by `(x+3)(x+2)` then we should get an equation that will allow us to solve for `A` and `B`.

`-> 5x+12 = A(x+2)+B(x+3)`

Now, to find `A` set `x=3` to cancel the second term and we get:

`5(-3)+12 = A(-3+2)+B(-3+3)`
`-3=-A -> A = 3`

Now set `x=-2` to obtain the value for for `B`.

`->5(-2) +12 = A(-2+2)+B(-2+3) -> B = 2`

So now we have that `A=3` and `B=2` we can re write the fraction given in the question as:

`(5x+12)/(x^2+5x+6) = 3/(x+3) + 2/(x+2)`

So we can now integrate:

`int3/(x+3) + 2/(x+2) dx = 3ln(x+3)+2ln(x+2) +C`