anonymous
  • anonymous
integral of (e^(3y))/(1+e^(6y))dy
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
This is what wolframalpha has: Take the integral: integral e^(3 x)/(1+e^(6 x)) dx For the integrand e^(3 x)/(1+e^(6 x)), substitute u = e^x and du = e^x dx: = integral u^2/(1+u^6) du For the integrand u^2/(1+u^6), substitute s = u^3 and ds = 3 u^2 du: = 1/3 integral 1/(1+s^2) ds The integral of 1/(1+s^2) is tan^(-1)(s): = 1/3 tan^(-1)(s)+constant Substitute back for s = u^3: = 1/3 tan^(-1)(u^3)+constant Substitute back for u = e^x: Answer: | | = 1/3 tan^(-1)(e^(3 x))+constant
anonymous
  • anonymous
What I don't get is why they get u^2 instead of u^3 for step 2.
anonymous
  • anonymous
No, for step two. Well, step one, really, i.e. the first time they use u substitution. The numerator originally equals e^(3x), and they let u=e^x. When they substitute back in, they get u^2 in the numerator, but it seems like it should be u^3.

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More answers

Jhannybean
  • Jhannybean
\[\large \int\limits \frac{e^{3y}}{1+e^{6y}}dy\] let u = 3y and du = 3dy -> du/3 =dy
Jhannybean
  • Jhannybean
\[\large \frac{1}{3}\int\limits \frac{e^u}{1+e^{2u}}du\]
Jhannybean
  • Jhannybean
I think that would be an easier way to solve this.
anonymous
  • anonymous
Where from there? U sub again?
anonymous
  • anonymous
That doesn't seem to work, I get stuck with an e^u that I can't do anything with.
Jhannybean
  • Jhannybean
Maybe.... you could integrate separately?... \[\large \frac{1}{3} [\int\limits e^udu*\int\limits \frac{1}{1+e^{2u}}du]\]
anonymous
  • anonymous
What's sad is that I just took calc 2.
anonymous
  • anonymous
Okay, yeah.
Jhannybean
  • Jhannybean
I just took calc 2 as well xD
anonymous
  • anonymous
I just have a crappy memory, sometimes the obvious isn't as obvious as it should be.
Jhannybean
  • Jhannybean
Do you know where to go from there?...
Jhannybean
  • Jhannybean
@Mertsj Need a little help simplifying :\
Jhannybean
  • Jhannybean
I'm just thinking, if we substituted 2u as something else, how would you show e^2? because 1/x^2+a^2 is tan inverse..
Jhannybean
  • Jhannybean
would \[\large \int\limits \frac{1}{1+e^m} = \frac{1}{1}\tan^{-1}(\frac{e^m}{1})\]
Jhannybean
  • Jhannybean
maybe....
anonymous
  • anonymous
that's what I'm wondering too, no root
Jhannybean
  • Jhannybean
i KNOW im on the right track, i've done this exact problem before, i'm missing a step so i know the conclusion,or what it's supposed to be somewhat... and the beginning step... but i don't quite know how to work out the rest. :\
anonymous
  • anonymous
can you make it a root?
Jhannybean
  • Jhannybean
there is no need to.
Jhannybean
  • Jhannybean
I'm just having trouble evaluating e^2u, that's all that's confusing me. once that's sorted out, i think the rest is understandable.
anonymous
  • anonymous
eg sqrt(1+e^(4u))
Mertsj
  • Mertsj
The reason it is not u^3 is because e^3y=e^2y(e^y) u=e^y and u^2=e^2y and du = e^y dy
Mertsj
  • Mertsj
So u^2 du=e^2y(e^y dy)=e^3ydy
Jhannybean
  • Jhannybean
yikes...
Jhannybean
  • Jhannybean
WAIT A MOMENT.
anonymous
  • anonymous
thanks for that mertsj
Mertsj
  • Mertsj
|dw:1370139872747:dw|
Mertsj
  • Mertsj
|dw:1370140062329:dw|
Jhannybean
  • Jhannybean
WHAT IF. \[\large \int\limits \frac{e^u}{1+e^{2u}}du \] let v = e^u dv =e^u du so... \[\large \int\limits \frac{v}{1+v^2}*\frac{dv}{v}\]\[\large \int\limits \frac{1}{1+v^2}dv\]\[\large \frac{1}{1}\tan^{-1}(\frac{v}{1})+c\]\[\large \tan^{-1}(v) +c\] substitute your v and u back in and simplify, you'll get your answer
Mertsj
  • Mertsj
Very good.
anonymous
  • anonymous
*applause*
anonymous
  • anonymous
One question: did the v/1 come from the dv?
Jhannybean
  • Jhannybean
no it came from the definition of tan^(-1)\[\large \int\limits \frac{1}{x^2+a^2} = \frac{1}{a}\tan^{-1}(\frac{x}{a})\] where x is our variable and a is our constant. In our equation we have \[\large \int\limits \frac{1}{1+v^2}dv\] where v = e^(2u) or u^2 so putting that in our equation, we would have \[\large \frac{1}{1}\tan^{-1}(\frac{v}{1})\]
Jhannybean
  • Jhannybean
see how it works?
anonymous
  • anonymous
Ahhh, ok. Thanks a lot jhannybean.
Jhannybean
  • Jhannybean
No problemo! and thanks to you too @Mertsj :)
Mertsj
  • Mertsj
yw

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