Hi,

http://i1104.photobucket.com/albums/...ralproblem.png

Can anyone explain how this follows? Has it been factories and simplified?

Thanks!

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- Jan 5th 2011, 07:28 AMAntIntegral of ratio
Hi,

http://i1104.photobucket.com/albums/...ralproblem.png

Can anyone explain how this follows? Has it been factories and simplified?

Thanks! - Jan 5th 2011, 07:33 AMChris L T521
- Jan 5th 2011, 07:36 AMchisigma
Is...

$\displaystyle \displaystyle \frac{2\ t^{3}}{1+t} = 2\ t^{2} - \frac{2\ t^{2}}{1+t} = 2\ t^{2} - 2\ t + \frac{2\ t}{1+t} = 2\ t^{2} - 2\ t + 2 - \frac{2}{1+t}$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Jan 5th 2011, 08:18 AMAnt
Yes thanks, it's all quite straight forward from there. I think I need to revisit algebraic long division though it's been about 3 years since I've used it!

RE the method used directly above, in general when given:

A/B = (C.B)/B - D/B = C - D/B ???

This bears some resemblance to completely the square? - Jan 5th 2011, 10:06 AMTheCoffeeMachine
Here is a trick: let $\displaystyle y = 1+t$ so that $\displaystyle y-1 = t$, then:

$\displaystyle \begin{aligned} & \frac{2t^3}{1+t} = \frac{2(y-1)^3}{y} = \frac{2 y^3-6 y^2+6 y-2}{y} = \\& 2y^2-6y+6-\frac{2}{y} = 2(1+t)^2-6(1+t)\\&+6-\frac{2}{1+t} = 2t^2-2 t+2-\frac{2}{t+1}.\end{aligned} $

Or, better yet, use the substitution as an integral substitution. - Jan 5th 2011, 11:34 AMChris L T521