Help with e^x intgral

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Hey, everyone

I am working on a calc problem, and I have no idea where to start. The integral is

e^x
------------- [division problem]
(25+e^2x)^4

Do I let my u equal to the 25+e^2x? or what...
then after that what do i do.

Thanks for all the help in advance.
-Eiano
 

Answers and Replies

GCT
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Do I let my u equal to the 25+e^2x? or what...
then after that what do i do.
here's an ingenious idea, try it out, yes try the substitution you mentioned and we'll see what magically happens.
 
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Anyone else?
(i KNOW my u can't be 25+e^2x, that's what th OR WHAT was for)
but HAHA, u were so funny with that response.. haha
 
LeonhardEuler
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Partial fraction decomposition? It looks like it'll take a long time, though.
 
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Eiano said:
Anyone else?
(i KNOW my u can't be 25+e^2x, that's what th OR WHAT was for)
but HAHA, u were so funny with that response.. haha
He was serious.. that's the solution.
 
GCT
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uhmm, actually I wasn't, he he o:)

partial fractions...hmm....sounds interesting, we'll have to try it out, although I've never done partial fractions with e^x,e^2x as a variable.

[tex]u=(25+e^{2x}),~du=2e^{2x}dx,~e^{x}= \sqrt{u-25}[/tex]

next substitute u=1/t

simplifying will give you soley a square root function in the denominator, convert this to a completed square, the rest should be easy

I'm sure there's a more elegant solution though

latex is ****ing up, I've actually typed the whole problem and solution out through latex, however for some reason it's picking up some old latex data from a couple of months ago :confused:
 
GCT
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alright finally got the latex to appear

[tex]I = \int \frac{e^{x}dx}{(25+e^{2x})^4}[/tex]

[tex]u=(25+e^{2x}),~du=2e^{2x}dx,~e^{x}= \sqrt{u-25}[/tex]

[tex]I=.5 \int \frac{du}{u^{4} \sqrt{u-25}} [/tex]

[tex]u=1/t,~du=-1/t^{2} dt[/tex]

[tex]I=-.5 \int \frac{dt}{ \sqrt{1/t^{2} - 25/t}}[/tex]

[tex] -.5 \int \frac{dt}{ \sqrt{(1/t -12.5)^{2} - 12.5^{2}}} [/tex]

[tex] (1/t -12.5) = 12.5sec \theta [/tex]

the rest should be easy

etc...

see any errors, please point them out, I'm guessing that there's probably a more elegant solution
 
Last edited:
HallsofIvy
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But e2x is (ex)2 so 25+ e2x= 25+ (ex)2. Try letting u= ex first, so that du= exdx. Now the integral is [tex]\int \frac{du}{25+u^2}[/tex] and that smells like an arctangent. Let u= 5 v so that du= 5 dv and 25+ u2= 25(1+ v2). Now the integral is [tex]\frac{5}{5}\int \frac{dv}{v^2+1}= arctan(v)+ C[/tex].
Of course, that's [tex]arctan(\frac{u}{5})+ C= arctan(\frac{e^x}{5})+ C[/tex]
 
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TD
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Unfortunately, there was a 4th power in the denominator... :wink:
 
GCT
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But e2x is (ex)2 so 25+ e2x= 25+ (ex)2. Try letting u= ex first, so that du= exdx. Now the integral is
I'm not quite sure what you're getting at there can you clarify? also note as TD said, there's a fourth power in the denominator.

I'm not able to see anything wrong with my solution.
 
GCT
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HallsofIvy, nevermind, I thought you were pointing out an error in my solution
 
TD
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If it weren't for the fourth power, HallsofIvy would've had an easy solution. He just must have missed it, I don't think he meant that yours was wrong. Without the fourth power, yours just seemed so long :smile:
 
GCT
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his method might lead to a partial fractions solution
 
LeonhardEuler
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GCT said:
[tex]I=.5 \int \frac{du}{u^{4} \sqrt{u-25}} [/tex]
[tex]u=1/t,~du=-1/t^{2} dt[/tex]
[tex]I=-.5 \int \frac{dt}{ \sqrt{1/t^{2} - 25/t}}[/tex]
I think I might have found an error here, GCT. Substituting gives:
[tex]I=.5 \int \frac{(-1/t^2)dt}{ (1/t)^4\sqrt{1/t - 25}}[/tex]
[tex]=-.5 \int \frac{dt}{ (1/t)^2\sqrt{1/t - 25}}[/tex]
[tex]=-.5 \int \frac{dt}{ \sqrt{1/t^5 - 25/t^4}}[/tex]
 
GCT
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I'm not quite sure what you did there, note that bringing in 1/t^2 within the square root will reduce it to 1/t within the square root.
 
LeonhardEuler
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Doesn't it go like this:
[tex]\sqrt{a}\sqrt{b}=\sqrt{ab}[/tex]
[tex]\frac{1} {t^2}\sqrt{x} = \sqrt{\frac{1} {t^4}}\sqrt{x}=\sqrt{\frac{x} {t^4}}[/tex]
 
GCT
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yeah, you're right
 
GCT
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alright, shame on me

What we can do is

[tex]I=.5 \int \frac{du}{u^{4} \sqrt{u-25}} [/tex]
[tex]u=25sec^{2} \theta ,~du=25tan \theta d \theta [/tex]
[tex]I= \frac{-1}{10(25^{3})} \int cos^{8} \theta d \theta [/tex]
which can be solved using standard procedure ("table integral")

hopefull I haven't goofed up this time, I'll be posting the full version later
 
Last edited:
LeonhardEuler
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Oh, man bad news, GCT. If [itex]u=\sec^2{\theta}[/itex] then
[tex]\frac{du}{d\theta}=\tan{\theta}\sec^2{\theta}[/tex]
you were probably thinking of
[tex]\int \sec^2{\theta}=\tan{\theta}[/tex]
It's Ok, we all have bad days!
 
GCT
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I think I've been drinking way too much these days, anyways thanks for pointing that out.

So the modification would result in
[tex]I= \frac{-1}{10(25^{3})} \int cos^{6} \theta d \theta [/tex]
...right? :wink:
 
GCT
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so, so far I have

[tex]I = \int \frac{e^{x}dx}{(25+e^{2x})^4}[/tex]


[tex]u=(25+e^{2x}),~du=2e^{2x}dx,~e^{x}= \sqrt{u-25}[/tex]

[tex]I=.5 \int \frac{du}{u^{4} \sqrt{u-25}} [/tex]

[tex]u=25sec^{2} \theta ,~du=50tan \theta sec^{2}\theta d \theta [/tex]

[tex] \frac{-1}{25^{2}10} \int cos^{6} \theta d \theta [/tex]

which can be solved easily using standard procedure (trignometric integrals, formula should be listed within the list of table integral-back of text)
 
Last edited:
LeonhardEuler
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Looks right to me.
 

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