1. ## Integral of ratio

Hi,

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

Thanks!

2. 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?

3. 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

$\chi$ $\sigma$

4. 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?

5. 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.

6. 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}