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

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

jotopia34
 one year 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

wio
 one year ago
Best ResponseYou've already chosen the best response.1Okay, what exactly is the equation?

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

wio
 one year ago
Best ResponseYou've already chosen the best response.1I do not know what the correct equation is.

jotopia34
 one year ago
Best ResponseYou've already chosen the best response.0the euation is above

wio
 one year ago
Best ResponseYou've already chosen the best response.1There are MANY equations above.

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

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

wio
 one year ago
Best ResponseYou've already chosen the best response.1\[ \Large \int_0^1 \frac{e^x}{e^{2x}+1} dx \]

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1think of \[\int\frac{x}{1+x^2}dx\]

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

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

jotopia34
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0thats right jotopia... thats y i knew i did it wrong XD

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1\[u=e^x,du=e^xdx\] you get \[\int\frac{du}{1+u^2}\]

jotopia34
 one year 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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1oh it is a definite integral

jotopia34
 one year ago
Best ResponseYou've already chosen the best response.0This is scandalous!!

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1\(u(0)=1,u(1)=e\) you get \[\int_1^e\frac{du}{1+u^2}\]

wio
 one year ago
Best ResponseYou've already chosen the best response.1Since \(u=e^{2x}=(e^x)^2 \implies \sqrt{u}=e^x\), @Goten77 's method isn't hopeless.

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1anti derivative of \(\frac{1}{1+u^2}\) is \(\tan^{1}(u)\)

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1yes it certainly is

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1\((e^x)^2=e^{2x}\)

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

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1and also you are right, \(\cos^2(x)=(\cos(x))^2\)

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1it is just more common to write \(f^2(x)\) than \((f(x))^2\)

jotopia34
 one year 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 \[e ^{2}x ^{2}\]

wio
 one year ago
Best ResponseYou've already chosen the best response.1You're confusing multiplication with exponents

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

jotopia34
 one year 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

wio
 one year ago
Best ResponseYou've already chosen the best response.1\((ex)^2 = e^2x^2\) but \((e^x)^2 =e^{2x}\)

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

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

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

jotopia34
 one year 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
 one year 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!!

wio
 one year ago
Best ResponseYou've already chosen the best response.1Anyway, 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

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

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

wio
 one year ago
Best ResponseYou've already chosen the best response.1That's the antiderivative, you have to put in the limits.

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

wio
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, though I'd like to see your work.

Hoa
 one year 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

Hoa
 one year 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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