# Indefinite Integral of (3x^2-10)/(x^2-4x+4) dx using PARTIAL FRACTION

1. ### Susie babe

3
1. The problem statement, all variables and given/known data

Integrate (3x^2-10)/(x^2-4x+4) dx using partial fractions.

2. Relevant equations

None

3. The attempt at a solution

I tried using A/(x-2) + B/(x-2)^2 but I didnt get a coeffecient of an x^2.

I've also tried using (Ax+B)/(x-2) + C/(x-2)^2

Though I honestly thought that the usual A/(x-2) + B/(x-2)^2 would work, How do I know which formula to use. I know you have to do long division if the numerator has an x to a larger value than that of the denominator. I know when you have something like: (x^2-1) or (x^3+2) in the denominator then you use the Ax+B, Cx+D formula.

2. ### eumyang

1,347
Re: Indefinite Integral of (3x^2-10)/(x^2-4x+4) dx using PARTIAL FRAC

The last thing that you said is correct...
$$\frac{A}{x - 2} + \frac{B}{(x - 2)^2}$$
... because the denominator is a linear factor squared. But before you try partial fractions, you have to use long division because the degrees of the numerator and denominator are the same.

3. ### sandy.bridge

795
Re: Indefinite Integral of (3x^2-10)/(x^2-4x+4) dx using PARTIAL FRAC

The degree of the numerator is equal to the degree of the denominator. Try long division before partial fraction decomposition.

4. ### Ray Vickson

6,178
Re: Indefinite Integral of (3x^2-10)/(x^2-4x+4) dx using PARTIAL FRAC

Re-write the numerator as
$$3x^2-10 = 3(x^2-4x+4)+12x - 22.$$

5. ### Susie babe

3
Ah, so if the degree of the numerator and that of the denominator are the same then you have to use long division, didnt know that. Thanks a lot guys it worked out well.

Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted