Question

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

Answer 1

The integral equals `3ln|x + 2| -2 ln|x + 1| + C`

Explanation:

Note that the factoring of `x^2 + 3x + 2` is `(x + 2)(x + 1)`. Therefore:

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

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

`Ax + A + Bx+ 2B = x - 1`

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

We now have a system of equations: `{(A+ B = 1), (A + 2B = -1):}`

We can readily solve through elimination (subtract the second equation from the first to get the following):

`-B = 2`

`B = -2`

It is now clear that `A - 2 = 1 -> A = 3`. The integral becomes

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

We can now easily integrate.

`I = 3ln|x + 2| -2 ln|x + 1| + C`

Hopefully this helps!