Integration help

  • Thread starter franky2727
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  • #1
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how do i integrate (1+Z)/(2-2Z)
 

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  • #2
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What have you tried so far?
 
  • #3
HallsofIvy
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[tex]\int\frac{1+Z}{2- 2Z}dZ= \frac{1}{2}\int\frac{1+Z}{1- Z}dZ[/tex]
Does that help?

and
[tex]\frac{1+ Z}{1- Z}= -1- ?[/tex]
 
  • #4
Dick
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Try a u substitution. What's the obvious one?
 
  • #5
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substitute 2-2Z for u? but then ive still got something over something else and still dont get what method to use :S
 
  • #6
Defennder
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As HallsOfIvy pointed out, you can just use polynomial long division to decompose the fraction into something more integrable.
 
  • #7
Dick
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substitute 2-2Z for u? but then ive still got something over something else and still dont get what method to use :S

If u=2-2Z then Z=(2-u)/2. Substitute that into 1+Z. Now it's all u's.
 
  • #8
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Dick said it all. You need to write Z in terms of u, then you need to find dz/du by differentiating that. You substitute everything in and integrate.
 
  • #9
HallsofIvy
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But you will still have to divide the fraction. Seems simpler just to reduce the original fraction.
 
  • #10
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If you factor out the 2, sub u = 1-z and z = 1-u, all the work has been done for you. Just seperate the fraction and integrate.
 
  • #11
HallsofIvy
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And, one more time, why not just "separate the fraction" (by which I assume you mean "divide") without the substitution?
 
  • #12
Air
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If you substitute [itex]u=2-2z[/itex], and you simplify it, you will end up with this at one point:

[itex]-\frac12 \int \left( \frac{2}{u} - \frac12 \right) \ \mathrm{d}u \implies \frac14 \int 1 \ \mathrm{d}u - \int \frac{1}{u} \ \mathrm{d}u[/itex]

Integrate this and remember to substitute [itex]u=2-2z[/itex] in the end to get the final answer.
 
  • #13
pps
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[tex] u = z-1 [/tex]
[tex] du = dz [/tex]
[tex] z = u + 1 [/tex]

[tex]\frac{1}{2} \int \frac{1+z}{1-z}dz =\frac{1}{-2} \int \frac{1+z}{z-1}dz = [/tex]

[tex]\frac{1}{-2} \left( \int \frac{dz}{z-1} + \int \frac{z}{z-1}dz \right) = [/tex]

[tex]\frac{1}{-2} \left( \int \frac{du}{u} + \int \frac{u + 1}{u} du \right) = [/tex]

[tex]\frac{1}{-2} \left( \int \frac{du}{u} + \int \frac{du}{u} + \int du \right) = [/tex]

Continue from this point.
 
  • #14
HallsofIvy
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And if you use the fact that [tex]\frac{1+z}{2-2z}= -\frac{1}{2}\left(1+ \frac{1}{1-z}\right)[/tex] you have
[tex]\int \frac{1+z}{2- 2z}dz= -\frac{1}{2}\int (1+ \frac{1}{z-1}dz= -\frac{1}{2}(z+ ln|z-1|)+ C[/tex]

That seems simpler to me.
 
  • #15
pps
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And if you use the fact that [tex]\frac{1+z}{2-2z}= -\frac{1}{2}\left(1+ \frac{1}{1-z}\right)[/tex] you have
[tex]\int \frac{1+z}{2- 2z}dz= -\frac{1}{2}\int (1+ \frac{1}{z-1}dz= -\frac{1}{2}(z+ ln|z-1|)+ C[/tex]

That seems simpler to me.

Your method is indeed much simpler. =)
 
  • #16
Dick
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I don't think it's MUCH simpler. It's six of one and half a dozen of the other. Neither one is all that complicated. If you ask me. And you don't have to.
 
  • #17
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And if you use the fact that [tex]\frac{1+z}{2-2z}= -\frac{1}{2}\left(1+ \frac{1}{1-z}\right)[/tex]

That seems simpler to me.

That's false though... Just check z=0 the left side gives 1/2, but the right side gives -(1/2)(2)=-1...
 
  • #18
Dick
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At this point I'm wondering who cares. franky2727 hasn't checked in since the initial post. I guess this thread is just getting a lot of attention because nothing else is going on.
 
  • #19
HallsofIvy
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That's false though... Just check z=0 the left side gives 1/2, but the right side gives -(1/2)(2)=-1...

Typo: I meant
[tex]-\frac{1}{2}\left(1+ \frac{2}{z-1}\right)[/tex]
 

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