jotopia34
Can anyone help me with the Integral of
e^x/(1+e^x) dx
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Goten77
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|dw:1359688972018:dw|
Goten77
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hmm
jotopia34
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\[\int\limits_{0}^{1}e^x/e ^{2x}+1\]
jotopia34
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sorry I made a mistake its e^2x
Goten77
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oh then
wio
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Try u sub
wio
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\(u=e^x+1\)
jotopia34
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but it I do u=e^2x, du= 2e^2x and I still can't get rid of the e^x
wio
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Okay, what exactly is the equation?
Goten77
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|dw:1359689045480:dw|
jotopia34
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nope. I still dont have it.....
wio
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What is the equation?
wio
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I do not know what the correct equation is.
jotopia34
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the euation is above
wio
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There are MANY equations above.
wio
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Which one is correct?
jotopia34
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\[\int\limits_{0}^{1}e^x/e ^{2x}+1\]
wio
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So it isn't this?\[\Large
\int_0^1 \frac{e^2}{e^{2x}+1} dx
\]
wio
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\[
\Large
\int_0^1 \frac{e^x}{e^{2x}+1} dx \]
jotopia34
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yes
anonymous
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think of
\[\int\frac{x}{1+x^2}dx\]
Goten77
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|dw:1359689203929:dw| eh i think im doing something wrong....
Goten77
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and then satellite makes it look ez <.<
jotopia34
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wait, I thought you can't have a u and an x in the equation
Goten77
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thats right jotopia... thats y i knew i did it wrong XD
anonymous
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\[u=e^x,du=e^xdx\] you get
\[\int\frac{du}{1+u^2}\]
jotopia34
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I don't see what sattelite said because how can e^2x be like x^2
anonymous
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oh it is a definite integral
jotopia34
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This is scandalous!!
anonymous
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\(u(0)=1,u(1)=e\) you get
\[\int_1^e\frac{du}{1+u^2}\]
wio
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Since \(u=e^{2x}=(e^x)^2 \implies \sqrt{u}=e^x\), @Goten77 's method isn't hopeless.
anonymous
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anti derivative of \(\frac{1}{1+u^2}\) is \(\tan^{-1}(u)\)
Goten77
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this is 1 of those like situations u gotta memorize.... its um
arc tan i believe? im not sure.......been a while since i done this stuff
jotopia34
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oh, e^2x is the same as saying e^x squared????/
anonymous
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yes it certainly is
jotopia34
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like how cos^2x is (cosx)^2
anonymous
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\((e^x)^2=e^{2x}\)
anonymous
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laws 'o exponents
jotopia34
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it amazes me that no one ever pointed that out before
jotopia34
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lol. Laws 'o fuh shizzle
anonymous
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and also you are right, \(\cos^2(x)=(\cos(x))^2\)
anonymous
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it is just more common to write \(f^2(x)\) than \((f(x))^2\)
jotopia34
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yes, but its confusing because if you said (e^x)^2, would it be like saying \[e ^{2}x ^{2}\]
wio
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You're confusing multiplication with exponents
Goten77
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jotopia look at this my pic below
jotopia34
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hmmm, but isnt (2x)^2 the same as 2^2x^2. And e is a number
wio
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\((ex)^2 = e^2x^2\) but \((e^x)^2 =e^{2x}\)
Goten77
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|dw:1359689988230:dw|
jotopia34
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oh yeah!! I get it now
jotopia34
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that is key. You are all fantabulous
wio
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Anyway, if you sub \(u = e^{2x}+1\) then \(e^x = \sqrt{u^2-1}\) which is also complicated
jotopia34
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damn, nope I am still confused because if i asked you to take the derivative of \[(e ^{x})^2\], you would do \[2(e^x)(e^x) = 2(e^x)^2=2e ^{2x}\]. Yep i get it
jotopia34
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damn!! I still don't get it
\[u=e^{2x}, du=2e^{2x}dx \]
=\[\int\limits_{0}^{1} \frac{ e^x}{ 1+u }* \frac{ du }{ 2u }\], lost!!
wio
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Anyway, satelitte73's answer is correct, but if you didn't memorize that \(d/dx \tan^-{1}(x)=1/(x^2+1)\) there is a way to do it with trig subs
jotopia34
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so, no u sub?
wio
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Well you two two substitutions:\[
1)\quad u= e^x \\
2)\quad u=\tan(\theta)
\]
jotopia34
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is the answer tan^-1e^x\[\tan^-1e^x\]
wio
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That's the anti-derivative, you have to put in the limits.
jotopia34
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yes you are right but is that the antiderivative part?
wio
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Yeah, though I'd like to see your work.
Hoa
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Hi friend, you have a formula for the problem, why don't you apply? let u =e^x ---> du = e^x dx. you have the new numerator is du, in denominator, you have u^2 +1 . and then change the interval : when x =0 , u =1, when x =1, u =e, so your new one is integral from 1 to e of du/u^2 + 1 = arctan (u) from 0 to 1 = arctan1 - arctan 0 . arctan 1 = pi/4 and arctan 0 = 0 . so the final result is pi/4.
is it familiar with you?
Hope this help
Hoa
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oh I'm sorry, arctan (e) - arctan 1. not as above. terribly sorry