# Integral with e

## Homework Statement

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

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.

Related Calculus and Beyond Homework Help News on Phys.org
Dick
Homework Helper
I like u=e^z better. That turns e^(2z) into u^2. Yes, that is part of the solution.

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)

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 :(

rock.freak667
Homework Helper
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 :(
$$\int \frac{e^z}{1+e^{2z}}dz$$

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?

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)?

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.

rock.freak667
Homework Helper
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)?
well yes...but arctan(ez)+C (never forget the constant of integration)

+C (never forget the constant of integration)
Costs me 1 point every time :grumpy: