Question

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

Answer 1

`int (x^2+x)/((x+2)(x-1)^2) dx =2/9 ln abs(x+2) + 7/9 ln abs(x-1) - 2/(3(x-1)) + C`

Explanation:

Since the denominator has already been factored for us, we can tell that we're looking for a partial fraction decomposition of the form:

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

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

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

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

Equating coefficients we get the following system of linear equations:

`{ (A+B=1), (-2A+B+C=1), (A-2B+2C=0) :}`

Adding all three equations together, we find:

`3C = 2`

So `color(blue)(C=2/3)`

Subtracting the second equation from the first, we have:

`3A-C = 0`

Hence `color(blue)(A = 2/9)`

Then from the first equation we find `color(blue)(B=7/9)`

So:

`(x^2+x)/((x+2)(x-1)^2)=2/(9(x+2))+7/(9(x-1))+2/(3(x-1)^2)`

Hence:

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

`=int 2/(9(x+2))+7/(9(x-1))+2/(3(x-1)^2) dx`

`=2/9 ln abs(x+2) + 7/9 ln abs(x-1) - 2/(3(x-1)) + C`