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integration

Started by neep, Apr 23 2009 04:04 PM

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#1 neep

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Posted 23 April 2009 - 04:04 PM

Could someone please help me to integrate (4z^2)/(z-1)

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#2 Marcus

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Posted 23 April 2009 - 08:13 PM

Right since this is calculus there are probably multiple ways to solve this...but this is the way i'd do it
first you need to simplify the fraction, you could do this by long division, but since that would be difficult to write out here, I'll do it a different way
$\frac{z^{2}}{z-1}$ (i'm missing out the four for now, to make things easier)
$\frac{z^{2}}{z-1}=\frac{z^{2}-z+z}{z-1}$ all i've done here is add and subtract z...which is legal as it doesn't change the value of the equation
$\frac{z(z-1)+z}{z-1} = z + \frac{z-1+1}{z-1} = z+1+\frac{1}{z-1}$ ...now we have the that fraction into something we can integrate so

$=\int{\frac{4z^{2}}{z-1}}dz \\ =4\int{z+1+\frac{1}{z-1}}dz \\ =4(\frac{1}{2}z^{2}+z+\ln{|z-1|}) \\ =2(z^{2}+2x+\ln{{|z-1|}^{2}})+c \\$

post back with anything you don't understand

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#3 neep

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Posted 23 April 2009 - 08:46 PM

:Thanks a lot marcus, thats so ridiculously simple its not obviouls lol. i kept trying to use integration using dummy functions and integration by parts what an idiot.

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#4 Marcus

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Posted 24 April 2009 - 08:32 PM

Yeah I know what you mean it can be sooo simple to try the wrong technique, generally if it is a polynomial over a polynomial try this long division/partial fractions technique to integrate

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