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jotopia34
 3 years ago
Can anyone help me with the Integral of
e^x/(1+e^x) dx
jotopia34
 3 years ago
Can anyone help me with the Integral of e^x/(1+e^x) dx

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jotopia34
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{1}e^x/e ^{2x}+1\]

jotopia34
 3 years ago
Best ResponseYou've already chosen the best response.0sorry I made a mistake its e^2x

jotopia34
 3 years ago
Best ResponseYou've already chosen the best response.0but it I do u=e^2x, du= 2e^2x and I still can't get rid of the e^x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay, what exactly is the equation?

jotopia34
 3 years ago
Best ResponseYou've already chosen the best response.0nope. I still dont have it.....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What is the equation?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I do not know what the correct equation is.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0There are MANY equations above.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Which one is correct?

jotopia34
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{1}e^x/e ^{2x}+1\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So it isn't this?\[\Large \int_0^1 \frac{e^2}{e^{2x}+1} dx \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \Large \int_0^1 \frac{e^x}{e^{2x}+1} dx \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0think of \[\int\frac{x}{1+x^2}dx\]

Goten77
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359689203929:dw eh i think im doing something wrong....

Goten77
 3 years ago
Best ResponseYou've already chosen the best response.0and then satellite makes it look ez <.<

jotopia34
 3 years ago
Best ResponseYou've already chosen the best response.0wait, I thought you can't have a u and an x in the equation

Goten77
 3 years ago
Best ResponseYou've already chosen the best response.0thats right jotopia... thats y i knew i did it wrong XD

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[u=e^x,du=e^xdx\] you get \[\int\frac{du}{1+u^2}\]

jotopia34
 3 years ago
Best ResponseYou've already chosen the best response.0I don't see what sattelite said because how can e^2x be like x^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh it is a definite integral

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\(u(0)=1,u(1)=e\) you get \[\int_1^e\frac{du}{1+u^2}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Since \(u=e^{2x}=(e^x)^2 \implies \sqrt{u}=e^x\), @Goten77 's method isn't hopeless.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0anti derivative of \(\frac{1}{1+u^2}\) is \(\tan^{1}(u)\)

Goten77
 3 years ago
Best ResponseYou've already chosen the best response.0this 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
 3 years ago
Best ResponseYou've already chosen the best response.0oh, e^2x is the same as saying e^x squared????/

jotopia34
 3 years ago
Best ResponseYou've already chosen the best response.0like how cos^2x is (cosx)^2

jotopia34
 3 years ago
Best ResponseYou've already chosen the best response.0it amazes me that no one ever pointed that out before

jotopia34
 3 years ago
Best ResponseYou've already chosen the best response.0lol. Laws 'o fuh shizzle

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and also you are right, \(\cos^2(x)=(\cos(x))^2\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it is just more common to write \(f^2(x)\) than \((f(x))^2\)

jotopia34
 3 years ago
Best ResponseYou've already chosen the best response.0yes, but its confusing because if you said (e^x)^2, would it be like saying \[e ^{2}x ^{2}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You're confusing multiplication with exponents

Goten77
 3 years ago
Best ResponseYou've already chosen the best response.0jotopia look at this my pic below

jotopia34
 3 years ago
Best ResponseYou've already chosen the best response.0hmmm, but isnt (2x)^2 the same as 2^2x^2. And e is a number

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\((ex)^2 = e^2x^2\) but \((e^x)^2 =e^{2x}\)

jotopia34
 3 years ago
Best ResponseYou've already chosen the best response.0oh yeah!! I get it now

jotopia34
 3 years ago
Best ResponseYou've already chosen the best response.0that is key. You are all fantabulous

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Anyway, if you sub \(u = e^{2x}+1\) then \(e^x = \sqrt{u^21}\) which is also complicated

jotopia34
 3 years ago
Best ResponseYou've already chosen the best response.0damn, 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
 3 years ago
Best ResponseYou've already chosen the best response.0damn!! 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!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Anyway, 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well you two two substitutions:\[ 1)\quad u= e^x \\ 2)\quad u=\tan(\theta) \]

jotopia34
 3 years ago
Best ResponseYou've already chosen the best response.0is the answer tan^1e^x\[\tan^1e^x\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's the antiderivative, you have to put in the limits.

jotopia34
 3 years ago
Best ResponseYou've already chosen the best response.0yes you are right but is that the antiderivative part?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, though I'd like to see your work.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hi 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh I'm sorry, arctan (e)  arctan 1. not as above. terribly sorry
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