How can I integrate (e^7x)/(e^(14x) + 9)dx using u-substitution?

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Homework Statement


So, I have to integrate the following expression:
(e^7x)/(e^(14x) + 9)dx


Homework Equations



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



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!
 
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Hint: e14x=(e7x)2
 
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
 
But you can use a subsitution: use, say, u=e^{7x}
 
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!
 
jrmed13 said:
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!

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

And yes, the thing being integrated is called the integrand :smile:
 
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!
 
jrmed13 said:
sorry for my constant refrain of complaining, but I already tried that and couldn't figure out how to integrate (u^2 + 9)^-1...

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

And what about the 9??
That derivative is for x^2 + 1, not x^2 + 9
 
  • #10
I mean that integral instead of derivative
 
  • #11
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
Glad I could help.

In case you're interested, the way to transform x^2+9 into the above form is to make a substitution of the form v=x/3, which will give 9(v^2-1). Then use v=tany So, you can solve this problem using around 4 different substitutions in succession.
 
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