Question

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

Answer 1

The answer is `==-1/(2(x-2)^2)-2/(x-2)+ ln(|x-2|)+C`

Explanation:

The numerator is

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

Let's perform the decomposition into partial fractions

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

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

The denominators are the same, we compare the numerators

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

Let `x=2`, `=>`, `1=A`

Coefficients of `x^2`

`1=C`

Coefficients of `x`

`-2=B-4C`

`B=4C-2=4-2=2`

Therefore,

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

So,

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

`=-1/(2(x-2)^2)-2/(x-2)+ ln(|x-2|)+C`

Answer 2

I got:

`ln|x-2| - 2/(x-2) - 1/(2(x-2)^2) + C`

(without partial fractions.)

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You don't have to use partial fractions... If you want to do it, you can see the other answer, but you don't have to do it unless you are asked to.

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

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

`= int (xcancel((x - 2)))/(x-2)^(cancel(3)^(2)) + 1/(x-2)^3dx`

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

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

These integrals are fundamental enough that partial fractions can be avoided. Always see if there is an easier way to do a math problem.

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

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