Integration help

1. Jul 30, 2008

franky2727

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

2. Jul 30, 2008

jeffreydk

What have you tried so far?

3. Jul 30, 2008

HallsofIvy

Staff Emeritus
$$\int\frac{1+Z}{2- 2Z}dZ= \frac{1}{2}\int\frac{1+Z}{1- Z}dZ$$
Does that help?

and
$$\frac{1+ Z}{1- Z}= -1- ?$$

4. Jul 30, 2008

Dick

Try a u substitution. What's the obvious one?

5. Jul 30, 2008

franky2727

substitute 2-2Z for u? but then ive still got something over something else and still dont get what method to use :S

6. Jul 30, 2008

Defennder

As HallsOfIvy pointed out, you can just use polynomial long division to decompose the fraction into something more integrable.

7. Jul 30, 2008

Dick

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

8. Jul 30, 2008

roam

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. Jul 31, 2008

HallsofIvy

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

10. Jul 31, 2008

moemoney

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. Jul 31, 2008

HallsofIvy

Staff Emeritus
And, one more time, why not just "separate the fraction" (by which I assume you mean "divide") without the substitution?

12. Jul 31, 2008

Air

If you substitute $u=2-2z$, and you simplify it, you will end up with this at one point:

$-\frac12 \int \left( \frac{2}{u} - \frac12 \right) \ \mathrm{d}u \implies \frac14 \int 1 \ \mathrm{d}u - \int \frac{1}{u} \ \mathrm{d}u$

Integrate this and remember to substitute $u=2-2z$ in the end to get the final answer.

13. Aug 1, 2008

pps

$$u = z-1$$
$$du = dz$$
$$z = u + 1$$

$$\frac{1}{2} \int \frac{1+z}{1-z}dz =\frac{1}{-2} \int \frac{1+z}{z-1}dz =$$

$$\frac{1}{-2} \left( \int \frac{dz}{z-1} + \int \frac{z}{z-1}dz \right) =$$

$$\frac{1}{-2} \left( \int \frac{du}{u} + \int \frac{u + 1}{u} du \right) =$$

$$\frac{1}{-2} \left( \int \frac{du}{u} + \int \frac{du}{u} + \int du \right) =$$

Continue from this point.

14. Aug 1, 2008

HallsofIvy

Staff Emeritus
And if you use the fact that $$\frac{1+z}{2-2z}= -\frac{1}{2}\left(1+ \frac{1}{1-z}\right)$$ you have
$$\int \frac{1+z}{2- 2z}dz= -\frac{1}{2}\int (1+ \frac{1}{z-1}dz= -\frac{1}{2}(z+ ln|z-1|)+ C$$

That seems simpler to me.

15. Aug 1, 2008

pps

Your method is indeed much simpler. =)

16. Aug 2, 2008

Dick

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. Aug 2, 2008

d_leet

That's false though... Just check z=0 the left side gives 1/2, but the right side gives -(1/2)(2)=-1...

18. Aug 2, 2008

Dick

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. Aug 2, 2008

HallsofIvy

Staff Emeritus
Typo: I meant
$$-\frac{1}{2}\left(1+ \frac{2}{z-1}\right)$$