Integral of ratio

**Ant** · January 5th 2011, 07:28 AM

Hi,

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

Thanks!

**Chris L T521** · January 5th 2011, 07:33 AM

Originally Posted by Ant

Hi,

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?

**chisigma** · January 5th 2011, 07:36 AM

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

$\chi$ $\sigma$

**Ant** · January 5th 2011, 08:18 AM

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?

**TheCoffeeMachine** · January 5th 2011, 10:06 AM

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.

**Chris L T521** · January 5th 2011, 11:34 AM

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,

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

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