# 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
$$-\frac{1}{2}\left(1+ \frac{2}{z-1}\right)$$