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how do i integrate (1+Z)/(2-2Z)

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- Thread starter franky2727
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- #1

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how do i integrate (1+Z)/(2-2Z)

- #2

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What have you tried so far?

- #3

HallsofIvy

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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?

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- #6

Defennder

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- #7

Dick

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If u=2-2Z then Z=(2-u)/2. Substitute that into 1+Z. Now it's all u's.

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- #9

HallsofIvy

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But you will still have to divide the fraction. Seems simpler just to reduce the original fraction.

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- #11

HallsofIvy

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[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

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[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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[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

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[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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- #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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- #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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