# Integration help

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

What have you tried so far?

HallsofIvy
Homework Helper
$$\int\frac{1+Z}{2- 2Z}dZ= \frac{1}{2}\int\frac{1+Z}{1- Z}dZ$$
Does that help?

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

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

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

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

Dick
Homework Helper
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.

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.

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

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.

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

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.

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

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

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

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

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

Continue from this point.

HallsofIvy
Homework Helper
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.

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.

Your method is indeed much simpler. =)

Dick
Homework Helper
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.

And if you use the fact that $$\frac{1+z}{2-2z}= -\frac{1}{2}\left(1+ \frac{1}{1-z}\right)$$

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...

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