• Support PF! Buy your school textbooks, materials and every day products Here!

Integral with e

  • #1

Homework Statement



[tex]\int[/tex][tex]\frac{e^{z}}{1+e^{2z}}[/tex]

This is clearly a u substitution problem. My instinct is to let u = [tex]{1+e^{2z}[/tex]
so du = 2e^[tex]{2z}[/tex]dz :confused:

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.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
I like u=e^z better. That turns e^(2z) into u^2. Yes, that is part of the solution.
 
  • #3
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
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
rock.freak667
Homework Helper
6,230
31
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 :(
[tex]\int \frac{e^z}{1+e^{2z}}dz[/tex]

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

so you will be left with

[tex]\int \frac{1}{1+u^2} du[/tex]

NOT

[tex]\int \frac{u}{1+u^2}du[/tex]

can you integrate 1/(1+u2) du?
 
  • #6
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
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
rock.freak667
Homework Helper
6,230
31
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)
 
  • #9
+C (never forget the constant of integration)
Costs me 1 point every time :grumpy:
 

Related Threads on Integral with e

  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
6
Views
794
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
1K
Replies
5
Views
2K
  • Last Post
Replies
4
Views
640
  • Last Post
Replies
4
Views
767
  • Last Post
Replies
13
Views
1K
Top