OpenStudy (anonymous):

integral of (e^(3y))/(1+e^(6y))dy

4 years ago
OpenStudy (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

4 years ago
OpenStudy (anonymous):

What I don't get is why they get u^2 instead of u^3 for step 2.

4 years ago
OpenStudy (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.

4 years ago
OpenStudy (jhannybean):

\[\large \int\limits \frac{e^{3y}}{1+e^{6y}}dy\] let u = 3y and du = 3dy -> du/3 =dy

4 years ago
OpenStudy (jhannybean):

\[\large \frac{1}{3}\int\limits \frac{e^u}{1+e^{2u}}du\]

4 years ago
OpenStudy (jhannybean):

I think that would be an easier way to solve this.

4 years ago
OpenStudy (anonymous):

Where from there? U sub again?

4 years ago
OpenStudy (anonymous):

That doesn't seem to work, I get stuck with an e^u that I can't do anything with.

4 years ago
OpenStudy (jhannybean):

Maybe.... you could integrate separately?... \[\large \frac{1}{3} [\int\limits e^udu*\int\limits \frac{1}{1+e^{2u}}du]\]

4 years ago
OpenStudy (anonymous):

What's sad is that I just took calc 2.

4 years ago
OpenStudy (anonymous):

Okay, yeah.

4 years ago
OpenStudy (jhannybean):

I just took calc 2 as well xD

4 years ago
OpenStudy (anonymous):

I just have a crappy memory, sometimes the obvious isn't as obvious as it should be.

4 years ago
OpenStudy (jhannybean):

Do you know where to go from there?...

4 years ago
OpenStudy (jhannybean):

@Mertsj Need a little help simplifying :\

4 years ago
OpenStudy (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..

4 years ago
OpenStudy (jhannybean):

would \[\large \int\limits \frac{1}{1+e^m} = \frac{1}{1}\tan^{-1}(\frac{e^m}{1})\]

4 years ago
OpenStudy (jhannybean):

maybe....

4 years ago
OpenStudy (anonymous):

that's what I'm wondering too, no root

4 years ago
OpenStudy (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. :\

4 years ago
OpenStudy (anonymous):

can you make it a root?

4 years ago
OpenStudy (jhannybean):

there is no need to.

4 years ago
OpenStudy (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.

4 years ago
OpenStudy (anonymous):

eg sqrt(1+e^(4u))

4 years ago
OpenStudy (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

4 years ago
OpenStudy (mertsj):

So u^2 du=e^2y(e^y dy)=e^3ydy

4 years ago
OpenStudy (jhannybean):

yikes...

4 years ago
OpenStudy (jhannybean):

WAIT A MOMENT.

4 years ago
OpenStudy (anonymous):

thanks for that mertsj

4 years ago
OpenStudy (mertsj):

|dw:1370139872747:dw|

4 years ago