Question

How do you find `int (3x^2 - 10) / (x^2-4x+4) dx` using partial fractions?

Answer 1

`3x+12lnabs(x-2)-2/(x-2)+C`

Explanation:

First, since the degrees of the numerator and denominator are equal, use polynomial long division to rewrite the expression:

`(3x^2-10)/(x^2-4x+4)=3+(12x-22)/(x^2-4x+4)`

Now, perform partial fraction decomposition on `(12x-22)/(x^2-4x+4)`, recognizing that `x^2-4x+4=(x-2)^2`.

`(12x-22)/((x-2)^2)=A/(x-2)+B/(x-2)^2`

Note that since the term is squared, it will be repeated.

Multiply both sides by `(x-2)^2` to see that

`12x-22=A(x-2)+B`

When we set `x=2`, we see that

`12(2)-22=A(0)+B`

`2=B`

Arbitrarily, set `x=3` to solve for `A`, recalling that `B=2`:

`12(3)-22=A(1)+2`

`A=12`

Thus,

`(3x^2-10)/(x^2-4x+4)=3+12/(x-2)+2/(x-2)^2`

Now, we can integrate more simply:

`int3+12/(x-2)+2/(x-2)^2dx`

`=3x+12lnabs(x-2)-2/(x-2)+C`