# Integral with e

1. Jun 10, 2009

### phantomcow2

1. The problem statement, all variables and given/known data

$$\int$$$$\frac{e^{z}}{1+e^{2z}}$$

This is clearly a u substitution problem. My instinct is to let u = $${1+e^{2z}$$
so du = 2e^$${2z}$$dz

Now what? I don't see how this can be substituted into the original integral.

I could let u = e^z, but then that leaves e^2z. Clearly du in this case be e^zdz, but that also doesn't substitute in. e^2z = e^z * e^z. Could this be part of the solution? If somebody could provide some hint to get me going, I'd be grateful.

2. Jun 10, 2009

### Dick

I like u=e^z better. That turns e^(2z) into u^2. Yes, that is part of the solution.

3. Jun 10, 2009

### Physics_Math

I agree. Try u=exp(z). du=exp(z) dz. Then you should get a very familiar integral. (The solution will be a trig function)

4. Jun 10, 2009

### phantomcow2

Okay, so if I use e^z = u and du = e^zdz, I am left with

u/(1+u^2)dz

This is for my calc 2 class. I have no idea how a trig function would tie into this though :(

5. Jun 10, 2009

### rock.freak667

$$\int \frac{e^z}{1+e^{2z}}dz$$

so yes, with u=ez => du=ez dz

so you will be left with

$$\int \frac{1}{1+u^2} du$$

NOT

$$\int \frac{u}{1+u^2}du$$

can you integrate 1/(1+u2) du?

6. Jun 10, 2009

### phantomcow2

Ah, you're right.

I took calculus 1 at my college, but I am taking calculus 2 at a different school. If I understand what an integral is, then we are seeking a function whose derivative IS 1/1+x^2. My calc 1 class never covered this, but I am beggining to suspect the calc 1 at this new college did: inverse trig functions. A peek in the back of my textbook indicates that 1/1+x^2 is the derivative of the arctan function. Is that it?

So I'm simply looking at arctan (e^z)?

7. Jun 10, 2009

### phantomcow2

I appreciate all of your help, by the way. I hope I don't come across as helpless or not wanting to do the work. I've gotten through most of this assignment but may have encountered something I haven't seen before.

8. Jun 10, 2009

### rock.freak667

well yes...but arctan(ez)+C (never forget the constant of integration)

9. Jun 10, 2009

### phantomcow2

Costs me 1 point every time :grumpy:

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