How do you integrate `int (x+4) / (x^2-x-2) dx` using partial fractions?

The trinomial `x^2 - x - 2` can be factored as `(x - 2)(x + 1)`.

`A/(x- 2) + B/( x+ 1) = (x + 4)/((x - 2)(x + 1))`

`A(x+ 1) + B(x- 2) = x + 4`

`Ax + A + Bx - 2B = x + 4`

`(A + B)x + (A - 2B) = x + 4`

Write a system of equations:

`{(A + B = 1), (A - 2B = 4):}`

`B = 1 - A -> A - 2(1 - A) = 4`

`A - 2 + 2A = 4`

`3A = 6`

`A = 2`

`B = 1 - A = 1 - 2 = -1`

The partial fraction decomposition is therefore:

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

This can be integrated as `2ln|x - 2| - ln|x + 1| + C`

Hopefully this helps!

I got the same answer in a bit of a different approach.

`2ln|x-2| - ln|x+1| + C`

Sometimes if you can do something to simplify the problem, try it.

As an alternative answer, consider the following trick:

`int (x+4)/(x^2 - x - 2)dx`

`= int (x+4)/((x - 2)(x+1))dx`

`= int (color(red)(x-2+6))/((x - 2)(x+1))dx`

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

Now partial fractions becomes a little easier. The second integrand, ignoring the `6` for now, can be set up as

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

so that we get upon cross-multiplying,

`A(x+1) + B(x-2) = 1`,

`=> (A + B)x + (A - 2B) = 1`,

thus giving the system of equations

`A + B = 0`,
`A - 2B = 1`.

Now we simply get

`A = -B`,
`-B - 2B = -3B = 1`,

so that `B = -1/3` and `A = 1/3` and the entire integral from the start of the problem becomes:

`color(blue)(int (x+4)/((x-2)(x+1))dx)`

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

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

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

`= color(blue)(2ln|x-2| - ln|x+1| + C)`