Question

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

Answer 1

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

Answer 2

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

Answer 3

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.

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

Where from there? U sub again?

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

Okay, yeah.

Answer 12

I just took calc 2 as well xD

Answer 13

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

Answer 14

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

Answer 15

@Mertsj Need a little help simplifying :\

Answer 16

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..

Answer 17

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

Answer 18

maybe....

Answer 19

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

Answer 20

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. :\

Answer 21

can you make it a root?

Answer 22

there is no need to.

Answer 23

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.

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

yikes...

Answer 28

WAIT A MOMENT.

Answer 29

thanks for that mertsj

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Answer 32

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Answer 37

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