Question

How do you use partial fraction decomposition to decompose the fraction to integrate `(x^2-x-8)/((x+1)(x^2+5x+6))`?

Answer 1

`(x^2-x-8)/((x+1)(x+2)(x+3))= -3/(x+1) + 2/(x+2) + 2/(x+3)`

Explanation:

`(x^2-x-8)/((x+1)(x^2+5x+6))`

First we need to finish factoring the denominator:

`(x^2-x-8)/((x+1)(x+2)(x+3))`

Now we want `A, B, "and " C` to get:

`A/(x+1) + B/(x+2) + C/(x+3) = (x^2-x-8)/((x+1)(x+2)(x+3))`

So we need:

`A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2) = x^2-x-8`

The left side can be rewritten:

`Ax^2+5Ax+6A+Bx^2+4Bx+3B+Cx^2+3Cx+2C`

and then:

`(A+B+C)x^2 +(5A+4B+3C)x+(6A+3B+2C)=x^2-x-8`

So we need to solve the system:

`A+B+C=1`

`5A+4B+3C=-1`

`6A+3B+2C=-8`

Multiplying the first equation by `-3` and adding to the second. Then the first times `-2` and add to the third, gets us:

`2A+B=-4`

`4A+B=-10`

So `2A = -6` and `A = -3`, which gets us `B = 2` and together these give us `C=2`.

`(x^2-x-8)/((x+1)(x+2)(x+3))= -3/(x+1) + 2/(x+2) + 2/(x+3)`