Question

u-substitution problem

Answer 1

i'll post it just a sec

Answer 2

\[\int\limits_{?}^{?}\frac{ e^2x }{ e^4x +36 }\] disregard the question marks, it's an indefinite integral

Answer 3

I feel like I'm pretty close to figuring it out, but might need a couple pointers. Should I set u=e^2x ?

Answer 4

is the x part of an exponent?

Answer 5

oh darn it, yes. Yeah, the numerator should be \[e^(2x)\]

Answer 6

e^(2x)

Answer 7

need to wrap it in {..} so it reads it as a ... hmm, aint got a word for it. unit?

Answer 8

\[e^{2x}\]tada!!

Answer 9

oh I see, haha thanks. Okay, so if I use u= e^(2x), then I end up getting \[\frac{ 1 }{ 2 }\int\limits_{?}^{?}\frac{ 1 }{ u^2 +36 }\] is that right?

Answer 10

lets try\[\int\frac{ e^{2x} }{ e^{4x} +36 }dx\]
\[u=e^{4x} +36\]
what is du/dx?

Answer 11

hmm, usingthe e^(2x) up top might prove useful too

Answer 12

if you set u as that, then the du/dx would be 4e^(4x)

Answer 13

yeah, lets see if i can verify your idea

u= e^2x

du = 2 e^2x dx

du/2 = e^2x dx looks fine by me

Answer 14

wouldn't it be du/2e^2x = dx?

Answer 15

it can be, either way, the e^2x cancels out

Answer 16

Yeah, I'm just confused about what to do once I get to the \[\frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ 1 }{ u^2 +36 }\] part

Answer 17

think of another substitution, maybe a trig?

Answer 18

there aren't any trig functions in this equation though

Answer 19

there wasnt any "u"s in it either, but that didnt stop us did it?

Answer 20

okay, I'm just not sure how to use trig, could you explain what you mean?

Answer 21

say; u = 6 tan(v) just to use a different variable substitution

what is du/dv?

Answer 22

6secx right?

Answer 23

sec^2 but yes

Answer 24

okay, then what?

Answer 25

u = 6 tan(v)

du = 6sec^2(v) dv

start replacing parts again :)

Answer 26

i just found the solution on yahoo answers, http://answers.yahoo.com/question/index?qid=20081001192720AAksnm9 but thanks for the help anyway, I'll give you best response

Answer 27

\[\frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ du }{ u^2 +36 }\]
\[\frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ 6sec^2~dv }{ (6tan(v))^2 +36 }\]
\[\frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ 6sec^2~dv }{ 36tan^2(v) +36 }\]
\[\frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ 6sec^2~dv }{ 36(tan^2(v) +1) }\]
\[\frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ 6sec^2~dv }{ 36sec^2 }\]
\[\frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ sec^2~dv }{ 6sec^2 }\]
\[\frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ 1 }{ 6 }dv\]

Answer 28

\[F(v) = \frac12v+C\]
which can either be used as is, or shuffle it back up thru the ranks

Answer 29

u = 6 tan(v)\[tan^{-1}(\frac{u}{6})=v\]\[F(u(v)) = \frac{1}{12}tan^{-1}(\frac{u}{6})+C\]
and e^2x = u; \[F(x(u(v))) = \frac{1}{12}tan^{-1}(\frac{1}{6}e^{2x})+C\]

http://www.wolframalpha.com/input/?i=integrate+e%5E%282x%29%2F%28e%5E%284x%29%2B36%29

makes me think the wolf rewrote it a different but equal way