How do you integrate (x^2-2x+1)/(x-2)^3x2−2x+1(x−2)3 using partial fractions?

2 Answers
Narad T.
Mar 14, 2017

The answer is ==-1/(2(x-2)^2)-2/(x-2)+ ln(|x-2|)+C==−12(x−2)2−2x−2+ln(|x−2|)+C

Explanation:

The numerator is

x^2-2x+1=(x-1)^2x2−2x+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)(x−1)2(x−2)3=A(x−2)3+B(x−2)2+Cx−2

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

Let x=2x=2, =>⇒, 1=A1=A

Coefficients of x^2x2

1=C1=C

Coefficients of xx

-2=B-4C−2=B−4C

B=4C-2=4-2=2B=4C−2=4−2=2

Therefore,

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

So,

int((x-1)^2dx)/(x-2)^3=intdx/(x-2)^3+2intdx/(x-2)^2+intdx/(x-2)∫(x−1)2dx(x−2)3=∫dx(x−2)3+2∫dx(x−2)2+∫dxx−2

=-1/(2(x-2)^2)-2/(x-2)+ ln(|x-2|)+C=−12(x−2)2−2x−2+ln(|x−2|)+C

Truong-Son N.
Mar 15, 2017

I got:

ln|x-2| - 2/(x-2) - 1/(2(x-2)^2) + Cln|x−2|−2x−2−12(x−2)2+C

(without partial fractions.)

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∫x2−2x+1(x−2)3dx

= int (x^2 - 2x)/(x-2)^3 + 1/(x-2)^3dx=∫x2−2x(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)

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