# Thread: Problems with Partial Fractions

1. ## Problems with Partial Fractions

Hi, I've been working on these integral evaluations and I'm stuck. Any help on any of them would be greatly appreciated, thanks in advance!

1. $\displaystyle \int\frac{2x^3-3x+2}{x^3-x}dx$

My attempt:

$\displaystyle 2x^3-3x+2=\frac{A}{x^2-1}+\frac{B}{x-1}+\frac{C}{x}$ (Tried breaking it down, couldn't get it to work )

2. $\displaystyle \int\frac{x^4+1}{x^3+x}dx$

My attempt:

long division to get $\displaystyle \int xdx-\int\frac{x^2+1}{x^3+x}dx$ (not sure how to show that in Latex)

Then $\displaystyle x^2+1=\frac{A}{X^2+1}+\frac{B}{x+1}+\frac{C}{x}$

Not sure where to go from there...I imagine that I'm missing the same idea in both of these, so any help would be appreciated. Thanks again!

2. For #1:

$\displaystyle x^{3}-x=x(x^{2}-1)=x(x+1)(x-1)$

$\displaystyle \frac{2x^{3}-2x+2}{x^{3}-x}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x-1}$

Multiply both sides of the equation by $\displaystyle x(x+1)(x-1)$ and go on from there.

For #2:

$\displaystyle x^{3}+x=x(x^{2}+1)$

$\displaystyle \frac{x^{4}+1}{x^{3}+x}=\frac{A}{x}+\frac{Bx+C}{x^ {2}+1}$

Multiply both sides of the equation by $\displaystyle x(x^{2}+1)$ and go on from there.

Edit: Don't forget to divide beforehead as in Plato's post; I forgot that part.

3. Here is a solution.

4. Originally Posted by Pinkk

For #2:

$\displaystyle x^{3}+x=x(x^{2}+1)$

$\displaystyle \frac{x^{4}+1}{x^{3}+x}=\frac{A}{x}+\frac{Bx+C}{x^ {2}+1}$

Multiply both sides of the equation by $\displaystyle x(x^{2}+1)$ and go on from there.
Your response for number two is clearly incorrect as multiplication on both sides by the cubic will equate a quartic on the left to a quadratic on the right. The way it should be done is as follows:

$\displaystyle \frac{x^4+1}{x(x^2+1)} \equiv x + \frac{1-x^2}{x(x^2+1)}$

then

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

yielding

$\displaystyle \frac{x^4+1}{x(x^2+1)} \equiv x + \frac{1}{x} - \frac{2x}{x^2 +1}$ .

5. Yes, that's before I realized the highest degree of the numerator was larger than the highest degree of the denominator.

6. I think I've got them both, thanks all!

7. Originally Posted by Pinkk
Yes, that's before I realized the highest degree of the numerator was larger than the highest degree of the denominator.
My apologies for not noticing your edit.