# Indefinite Integral Question; e^x

1. Mar 3, 2012

1. The problem statement, all variables and given/known data
Find the indefinite integral (preferably using u-substitution):
∫(ex-e-x)2 dx

2. Relevant equations
N/A

3. The attempt at a solution
To be honest, I'm slightly confused as to which path I'm supposed to take with this, especially since I'm not sure what I should be using u to substitute here. This is one of my attempts at solving the question:

∫[(ex-e-x)2 dx]
= ∫[e2x dx] - ∫[e-2x dx]
= (e2x / 2) + (1 / 2e2x) + C

Could someone please tell me if I have the right answer / I'm on the right track / I'm completely wrong? If the latter, I'd appreciate it if you could help get me on track!

2. Mar 3, 2012

### BruceW

You almost got it. But you've forgotten the cross-terms when you did the square.

There is a good substitution you could use instead. But your method is just as good.

Edit: p.s. welcome to physicsforums, your name is good. reminds me of the town in pokemon

3. Mar 3, 2012

### rock.freak667

If you need the u-substitution, try just putting u=ex.

(Remember that 1/a is the same as a-1)

EDIT: Also in your attempt you should also have

∫e2x dx + ∫e-2x dx +∫2 dx

4. Mar 3, 2012

Before I continue asking questions, thanks for all the help! I really do appreciate it and I'm sorry if I'm a bit slow, haha.

Now then...
@BruceW: Thanks for the help, the welcome, and the name compliment! I'm actually a fan of the colour itself and the city in Pokemon, heh.
@rock.freak667: Ugh, how could I forget 1/a = a-1? Thanks for the reminder. I have a question about the +∫2 dx, though. How exactly did it get there?

I think I'm actually a bit more confused than I initially thought I was. Let's say I went with this method:

u=ex
du=exdx ; dx=du/ex

...and then what? This is what I'm leaning towards:
=∫(u2 dx) - ∫(1/u2 dx)

Do I substitute dx for du/ex or am I over-complicating things? Oh man, my head's gonna burst. @@

5. Mar 3, 2012

### SammyS

Staff Emeritus
(ex-e-x)2 = (ex-e-x)(ex-e-x)

Multiply it out. (Use FOIL, if that's what you're used to.)

6. Mar 3, 2012

### Dick

(e^x-e^(-x))^2=(e^x-e^(-x))*(e^x-e^(-x)). Multiply that out and you'll see where the '2' came from. rock.freak667 had the sign wrong on it.

7. Mar 3, 2012

Okay, do I tried expanding it and then went through this:

(ex-e-x)2 = (ex-e-x)(ex-e-x) = e2x + e-2x

∫(e2x + e-2x) dx
u = e2x
du = (e2x / 2) dx

Final solution:
(u2/e2x) + (2 / e2x)ln|e2x|

Last edited: Mar 3, 2012
8. Mar 3, 2012

### Staff: Mentor

No, this is incorrect, as Sammy and Dick pointed out. (a - b)2 ≠ a2 - b2, which is essentially what you are doing.
When you expand (ex - e-x)2 you should get three terms, not two.

Also, there is an error below - you shouldn't get any log terms.

9. Mar 3, 2012

Oh I see it now.
(ex - e-x)(ex - e-x) = e2x - 2 + e-2x

I somehow wrote "+ e0 - e0" at some point, oops. Thanks!
Time to try again. (:

10. Mar 3, 2012

### Staff: Mentor

For you last error, I think this is what you did:

∫dx/e2x = ln|e2x|

That is incorrect. Don't write the integrand as a fraction - use the negative exponents.

11. Mar 3, 2012

Okay, another attempt. Hope I fixed more than I broke, this time. :/
Thanks again everyone, I'm sorry for bothering you all. D:

∫(ex-e-x)2dx = ∫(e2x-2+e-2x)dx
= ∫e2xdx + ∫e-2xdx - ∫2dx

Let u = ex
du = exdx

∫e2xdx + ∫e-2xdx - ∫2dx = (1/ex)(u3/3)+(1/ex)(u-1/-1)-2x+C
= (1/ex)[(u3/3)+(u-1/-1)]-2x+C
= (1/ex)[(e3x/3) - (1/ex)]-2x+C
= (1/ex)[(e4x-3)/(3ex)]-2x+C
= [(e4x-3)/(3e2x)]-2x+C

Last edited: Mar 3, 2012
12. Mar 3, 2012

### SammyS

Staff Emeritus
What is $\displaystyle\int du \,?$

13. Mar 3, 2012

ex+C?

14. Mar 3, 2012

### SammyS

Staff Emeritus
Ignore my last post.

If u = ex and du = exdx,

then $\displaystyle\int e^{2x}dx$ becomes what integral in terms of u ?

15. Mar 3, 2012

∫u2dx? :x

[EDIT] Another attempt:
2e2x-2x-2e-2x+C

x_x

Last edited: Mar 3, 2012
16. Mar 3, 2012

### SammyS

Staff Emeritus
Yes. You appear to be confused.

You can't take 1/ex out of the integral. In fact you should leave it inside the integral as 1/u .

What is u2/u ?

17. Mar 3, 2012

u1 or just u? xD

(also, I edited my previous post with another answer in case you missed it.)

18. Mar 3, 2012

### SammyS

Staff Emeritus
OK, so what is $\displaystyle\int u \,du \,?$

19. Mar 3, 2012