Solving Integral with e: Tips and Tricks

[phantomcow2]

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.

 

[Dick]

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

 

[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)

 

[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 :(

 

[rock.freak667]

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=e^z => du=e^z 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?

 

[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)?

 

[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.

 

[rock.freak667]

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(e^z)+C (never forget the constant of integration)

 

[phantomcow2]

+C (never forget the constant of integration)

Costs me 1 point every time