# Thread: Partial fraction method on this problem

1. ## Partial fraction method on this problem

I'm just not getting this at all, and I lost the page from class where he did it.

I asked before and didn't get much further. I know there are a few methods for doing this.

The initial problem is the integral of dx/(x^2-1)^2 , I'm supposed to integrate by using partial fractions. I get to the point where, if I label my unknowns A, B, C, and D, I can figure out B and D are 1/4. Then I get stuck.

What's the easiest way to tackle this?

Thanks!

2. P.S, the steps he gave us were these:

1For f(x)=p(x)/q(x)

1. if degree P >= degree Q, divide it out.

2. Factor the denominator completely

3. Write the quotient as a sum.

4. Set the original fraction equal to the sum of the fraction.

5. Clear the equation of fractions.

6. Combine like terms.

7. Equate coefficients of like powers of X (? How do I do this?)

8. Solve for coefficients

3. Originally Posted by Wolvenmoon
P.S, the steps he gave us were these:

1For f(x)=p(x)/q(x)

1. if degree P >= degree Q, divide it out.

2. Factor the denominator completely

3. Write the quotient as a sum.

4. Set the original fraction equal to the sum of the fraction.

5. Clear the equation of fractions.

6. Combine like terms.

7. Equate coefficients of like powers of X (? How do I do this?)

8. Solve for coefficients
Ugh this took me forever!

You get:

$\displaystyle \displaystyle\frac{1}{(x^2-1)^2}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x-1}+\frac{D}{(x-1)^2}$

Which then gives:

$\displaystyle A(x+1)(x-1)^2+B(x-1)^2+C(x+1)^2(x-1)+D(x+1)^2=1$
So, you let $\displaystyle x=1$

$\displaystyle 4D=1$

$\displaystyle D=0.25$

So you're fine so far.

You let $\displaystyle x=-1$

$\displaystyle 4B=1$

$\displaystyle B=0.25$

Again, no problem.

But there isn't a substitution that will work for C and A.

So it gets a little more tricky.

Substitute in the values for B and D:

$\displaystyle A(x+1)(x-1)^2+\frac{1}{4}(x-1)^2+C(x+1)^2(x-1)+\frac{1}{4}(x+1)^2=1$

Now, the left hand side has to be equal to the right hand side, obviously.Read this paragraph slowly! There is $\displaystyle 0x^3$ on the right hand side of the equation. So there has to be a total of $\displaystyle 0x^3$ on the left. So, expand the brackets, but only look at the $\displaystyle x^3$ terms:
You'll get the following terms involving an $\displaystyle x^3$:

$\displaystyle Ax^3 + Cx^3$
So $\displaystyle Ax^3 + Cx^3 = 0x^3$ as there are no $\displaystyle x^3$ terms on the right.
Therefore 1) A + C = 0

Now I don't want to look at the $\displaystyle x^2$ terms because they will be tricky. I'll skip to terms which have no $\displaystyle x$ coefficients.

From the left hand side:

$\displaystyle A+\frac{1}{4}-C+\frac{1}{4}=1$
So 2) A-C=0.5

Two equations, two unknowns with A and C left to find.

Does that make sense?

4. Thank you!