# Polynomial Long Division

1. Nov 2, 2011

### Lancelot59

As part of solving a DE I need to make this friendlier to integrate:

$$\frac{2u}{1+u^{-2}}$$
I figured trying to divide it couldnt hurt. I got:
$$u^{-1}+u^{-3}$$

I can't type out all the steps easily, I'm on a mobile device at the moment. That answer looks suspect, did I do it correctly?

2. Nov 2, 2011

### Dick

It's more than suspect, it's completely wrong. If u=1 then your first expression is 1, the second one is 2. Pretty bad, yes? Suggest you get off the mobile device and type in your steps.

3. Nov 3, 2011

### symbolipoint

A more comfortable form of $\frac{2u}{1+u^{-2}}$ may be found if you start like this:
$\frac{2u}{1+u^{-2}} \frac{u^2}{u^2}$

4. Nov 3, 2011

### SteamKing

Staff Emeritus
Hmm... I've got 1 over some function times what is almost the derivative of that function ... and I want to integrate that combination ...

5. Nov 3, 2011

### Lancelot59

I'm home now.

I set it up like so (I left the two out):
$$u|1+u^{-2}$$

from my understanding you're supposed to write the functions in descending order of powers.

For 1:$$u*\frac{1}{u}=1$$
subtraction yields zero. You then need to find out what multiple of u gives you u-2. u*u-3=u-2, subtraction gives zero. Total remainder of zero.

So I get the final answer of:
$$\frac{1}{u}+u^{-3}+\frac{0}{1+u^{-2}}=\frac{1}{u}+u^{-3}$$

I think I just saw where I went wrong...I wrote it backwards.

I should be solving:
$$1+u^{-2}|u$$

I don't see how that works, it's missing an exponent.

6. Nov 3, 2011

### Dick

Right, you did it backwards. (1+u^(-2))/u=u^(-1)+u^(-3). The whole process is a lot easier to follow if you clear out the negative powers first, like symbolpoint suggested above.

7. Nov 3, 2011

### Lancelot59

Sounds like a plan! Thanks for the help.