# Integral of ratio

• January 5th 2011, 07:28 AM
Ant
Integral of ratio
Hi,

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

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

Thanks!
• January 5th 2011, 07:33 AM
Chris L T521
Quote:

Originally Posted by Ant
Hi,

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

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

Thanks!

Since you're integrating an improper fraction, we need to first make it proper.

By long division, you should see that $\dfrac{2t^3}{1+t}=2t^2-2t+2-\dfrac{2}{t+1}$

Now integrate this result instead. Can you proceed?
• January 5th 2011, 07:36 AM
chisigma
Is...

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

Kind regards

$\chi$ $\sigma$
• January 5th 2011, 08:18 AM
Ant
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?
• January 5th 2011, 10:06 AM
TheCoffeeMachine
Here is a trick: let $y = 1+t$ so that $y-1 = t$, then:

\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.
• January 5th 2011, 11:34 AM
Chris L T521
Quote:

Originally Posted by TheCoffeeMachine
Here is a trick: let $y = 1+t$ so that $y-1 = t$, then:

\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.

Or even better,

\begin{aligned}\dfrac{2t^3}{t+1} &= \dfrac{2(t^3+1)}{t+1}-\dfrac{2}{t+1}\\ &= \dfrac{2(t+1)(t^2-t+1)}{t+1}-\dfrac{2}{t+1}\\ &= 2t^2-2t+2-\dfrac{2}{t+1}\end{aligned}