# Integrals involving 'e'

1. Sep 6, 2008

### jrmed13

1. The problem statement, all variables and given/known data
So, I have to integrate the following expression:
(e^7x)/(e^(14x) + 9)dx

2. Relevant equations

We are doing the section on 'integrating by u-substitution' right now, so that might help in finding a solution...

3. The attempt at a solution
So, I tried a bunch of stuff - I tried u=7x, u=e^7x, u=14x, u=e^14x, u=x, etc. etc. and I can't get the right answer!!! I have been working on this problem for over an hour and am on the verge of tears!

2. Sep 6, 2008

### cristo

Staff Emeritus
Hint: e14x=(e7x)2

3. Sep 6, 2008

### jrmed13

I know that.. but how do I work from there???
The problem with "e" is that its derivative is different than integrating something like x^2

4. Sep 6, 2008

### cristo

Staff Emeritus
But you can use a subsitution: use, say, u=e^{7x}

5. Sep 6, 2008

### jrmed13

How??
I tried that million times and I couldn't get the right answer!!
Here's what I tried

u=e^7x
du=(7e^7x)dx

so then the integrand (that's what the thing being integrated is called, right?)
becomes
(u/(u^2 + 9))*du/(7e^7x)

that doesn't work!!!!!

6. Sep 6, 2008

### cristo

Staff Emeritus
Can you simplify the line in bold (perhaps using (#)?)

And yes, the thing being integrated is called the integrand

7. Sep 6, 2008

### jrmed13

sorry for my constant refrain of complaining, but I already tried that and couldn't figure out how to integrate (u^2 + 9)^-1........

I know it involves something with ln(u^2 + 9), but I couldn't get that to work either!

8. Sep 6, 2008

### cristo

Staff Emeritus
Oh, ok, well I can see why you're banging your head against a brick wall then! Integrals like $$\int\frac{dx}{1+x^2}$$ are ones you should really know (especially for exams etc..) but in essence most people just look them up. The way to solve this, however, would be to make a final substitution x=tany.

9. Sep 6, 2008

### jrmed13

Wait, what???
I don't follow - I have never seen that before in Calculus I (I'm in Calc. II right now)

That derivative is for x^2 + 1, not x^2 + 9

10. Sep 6, 2008

### jrmed13

I mean that integral instead of derivative

11. Sep 6, 2008

### jrmed13

Oh waiiiiiit never mind I found it in my reference page (integral of du/(x^2 + a^2) = blahblah.

Cristo, thank you so much - you are amaaaaaaaaaazing and although I did cry over this problem for about ninety minutes, I have learned a bunch and I am going to memorize some of these integrals involving u^2-a^2 and u^2+a^2, etc. that we didn't cover in AP Calc with flashcards.

Thanks again!!

12. Sep 6, 2008

### cristo

Staff Emeritus