1. Rational Function

Hi guys

$\displaystyle R(x)=6/(x^4-x)$

How do I use the mehod of partial fractions???? I'm rubbish with that method!

/Janus

PS:
$\displaystyle R(x)=7/(x+2)(x-1)$

A bonus Q :P Think I've got that A, B figured. but I really can't integrate it! *crikey!*... Gotta use the Linear Denominator, right?

2. Originally Posted by American-Pi
Hi guys

$\displaystyle R(x)=6/(x^4-x)$

How do I use the mehod of partial fractions???? I'm rubbish with that method!

/Janus

PS:
$\displaystyle R(x)=7/(x+2)(x-1)$

A bonus Q :P Think I've got that A, B figured. but I really can't integrate it! *crikey!*... Gotta use the Linear Denominator, right?
$\displaystyle \frac{6}{x^4-x} = \frac{6}{x(x^3 - 1)}$

If you know the formula for the difference of two cubes, then you can factorise $\displaystyle x^3 - 1^3$ even further.

Then you'll have something like this:

$\displaystyle \frac{6}{x(x+a)(x^2+bx+c} = \frac{A}{x} + \frac{B}{x+a} + \frac{Cx+D}{x^2 + bx + c}$

Which, if you multiply through by the denominator of the LHS, becomes:

$\displaystyle A(x+a)(x^2 + bx + c) + Bx(x^2 + bx + c) + (Cx+D)(x+a)x = 6$

Which you should be able to solve for A, B, C and D by expanding the LHS, and comparing coefficients of the RHS and LHS.

3. Thanks Champ!

4. Originally Posted by American-Pi
Thanks Champ!
Well that's even better!

$\displaystyle \frac{6}{x^3 - x} = \frac{6}{x(x^2 - 1)} = \frac{6}{x(x+1)(x-1)}$

$\displaystyle \therefore \frac{6}{x^3 - x} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{x-1}$

Multiply through by the denominator:

$\displaystyle 6 = A(x+1)(x-1) + Bx(x-1) + Cx(x+1)$

Now, if you let x = 0, you should be able to solve for A. If you let x = 1, you should be able to solve for C. If you let x = -1, you should be able to solve for B. Then you can plug those values into the partial fractions and integrate them with relative ease.

5. Find A and B by subtracting?

6. SORRY ABOUT 2x POST! Can't edit my old post!

$\displaystyle (A(x^2-2x+1)+B(x^2-x)+Cx)/x(x-1)^2$

???

7. Nope that seems wrong!

But I got a problem a lot like this Help any1?

8. Originally Posted by American-Pi
SORRY ABOUT 2x POST! Can't edit my old post!

$\displaystyle (A(x^2-2x+1)+B(x^2-x)+Cx)/x(x-1)^2$

???
Hmm, I'm not sure what you're trying to do here. I'll run you through it again with more documentation:

Here is the problem:

$\displaystyle \displaystyle \int \frac{6}{x^3 - x} \, dx$

The first thing I'll do is take out the integrand, and manipulate it separately to get it into a form that is more easily integrated.

$\displaystyle \frac{6}{x^3 - x}$

The first step is to take out a factor of x on the denominator:

$\displaystyle \frac{6}{x(x^2 - 1)}$

Now, $\displaystyle x^2 - 1$ is a difference of two squares. We know that if we have $\displaystyle x^2 - a^2$, then we can write that as $\displaystyle (x-a)(x+a)$. So $\displaystyle x^2 - 1^2 = (x-1)(x+1)$. So if we put that into our expression, we get the following:

$\displaystyle \frac{6}{x(x-1)(x+1)}$

Now, we must use partial fractions on that expression. That is, for every factorised term in the denominator, we want it to have it's own fraction separate from the rest of the denominator. So we'll have three fractions in total, we know what we want the denominators to be, we just don't know what the numerator will be... so we'll set our expression equal to the form that we want it to be in:

$\displaystyle \frac{6}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}$

Now, we need to find out what A, B and C are, and to do that, we need to use a few common 'tricks' used in partial fractions. The first thing we do is multiply the WHOLE equation by the denominator of the left hand side. So:

$\displaystyle \frac{6x(x-1)(x+1)}{x(x-1)(x+1)} =\frac{Ax(x-1)(x+1)}{x} + \frac{Bx(x-1)(x+1)}{x-1} + \frac{Cx(x-1)(x+1)}{x+1}$

Now we do some cancelling, and see what we're left with:

$\displaystyle 6 = A(x+1)(x-1) + Bx(x+1) + Cx(x-1)$

Now, from here it's rather easy to solve for A, B and C. We do it as follows:

Let $\displaystyle x = 0$

$\displaystyle 6 = A(1)(1) + B(0)(1) + C(0)(-1)$

$\displaystyle 6 = A$ ******

Let $\displaystyle x = 1$

$\displaystyle 6 = A(2)(0) + B(1)(2) + C(1)(0)$

$\displaystyle 2B = 6$ ******

Let $\displaystyle x = -1$

$\displaystyle 6 = A(0)(-2) + B(-1)(0) + C(-1)(-2)$

$\displaystyle 2C = 6$ ******

Using the equations that I've highlighted with stars, we can easily solve for A, B and C. Once we have A, B and C, we should insert them into this:

$\displaystyle \frac{6}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}$

And thus we will have re-written expression in a form that is easily to integrate. So our integral is now:

$\displaystyle \displaystyle \int \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1} \, dx$

This can be written:

$\displaystyle \displaystyle A \int \frac{dx}{x} + B \int \frac{dx}{x-1} + C \int \frac{dx}{x+1}$

To integrate this, you must simply remember that:

$\displaystyle \displaystyle \int \frac{dx}{x+a} = \ln(x+a) + C$ for any constant $\displaystyle a$.

I hope this helps you.