# Polynomial Long Division

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?

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Dick
Homework Helper
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.

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

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

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.
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:
$$1+u^{-2}|u$$

Hmm... I've got 1 over some function times what is almost the derivative of that function ... and I want to integrate that combination ...
I don't see how that works, it's missing an exponent.

Dick
Homework Helper
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.
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.

Sounds like a plan! Thanks for the help.