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jotopia34Best ResponseYou've already chosen the best response.0
\[\int\limits_{0}^{1}e^x/e ^{2x}+1\]
 one year ago

jotopia34Best ResponseYou've already chosen the best response.0
sorry I made a mistake its e^2x
 one year ago

jotopia34Best ResponseYou've already chosen the best response.0
but it I do u=e^2x, du= 2e^2x and I still can't get rid of the e^x
 one year ago

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

jotopia34Best ResponseYou've already chosen the best response.0
nope. I still dont have it.....
 one year ago

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

jotopia34Best ResponseYou've already chosen the best response.0
the euation is above
 one year ago

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

jotopia34Best ResponseYou've already chosen the best response.0
\[\int\limits_{0}^{1}e^x/e ^{2x}+1\]
 one year ago

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

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

satellite73Best ResponseYou've already chosen the best response.1
think of \[\int\frac{x}{1+x^2}dx\]
 one year ago

Goten77Best ResponseYou've already chosen the best response.0
dw:1359689203929:dw eh i think im doing something wrong....
 one year ago

Goten77Best ResponseYou've already chosen the best response.0
and then satellite makes it look ez <.<
 one year ago

jotopia34Best ResponseYou've already chosen the best response.0
wait, I thought you can't have a u and an x in the equation
 one year ago

Goten77Best ResponseYou've already chosen the best response.0
thats right jotopia... thats y i knew i did it wrong XD
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
\[u=e^x,du=e^xdx\] you get \[\int\frac{du}{1+u^2}\]
 one year ago

jotopia34Best ResponseYou've already chosen the best response.0
I don't see what sattelite said because how can e^2x be like x^2
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
oh it is a definite integral
 one year ago

jotopia34Best ResponseYou've already chosen the best response.0
This is scandalous!!
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
\(u(0)=1,u(1)=e\) you get \[\int_1^e\frac{du}{1+u^2}\]
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.1
anti derivative of \(\frac{1}{1+u^2}\) is \(\tan^{1}(u)\)
 one year ago

Goten77Best ResponseYou've already chosen the best response.0
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
 one year ago

jotopia34Best ResponseYou've already chosen the best response.0
oh, e^2x is the same as saying e^x squared????/
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
yes it certainly is
 one year ago

jotopia34Best ResponseYou've already chosen the best response.0
like how cos^2x is (cosx)^2
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
\((e^x)^2=e^{2x}\)
 one year ago

jotopia34Best ResponseYou've already chosen the best response.0
it amazes me that no one ever pointed that out before
 one year ago

jotopia34Best ResponseYou've already chosen the best response.0
lol. Laws 'o fuh shizzle
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
and also you are right, \(\cos^2(x)=(\cos(x))^2\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
it is just more common to write \(f^2(x)\) than \((f(x))^2\)
 one year ago

jotopia34Best ResponseYou've already chosen the best response.0
yes, but its confusing because if you said (e^x)^2, would it be like saying \[e ^{2}x ^{2}\]
 one year ago

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

Goten77Best ResponseYou've already chosen the best response.0
jotopia look at this my pic below
 one year ago

jotopia34Best ResponseYou've already chosen the best response.0
hmmm, but isnt (2x)^2 the same as 2^2x^2. And e is a number
 one year ago

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

jotopia34Best ResponseYou've already chosen the best response.0
oh yeah!! I get it now
 one year ago

jotopia34Best ResponseYou've already chosen the best response.0
that is key. You are all fantabulous
 one year ago

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

jotopia34Best ResponseYou've already chosen the best response.0
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
 one year ago

jotopia34Best ResponseYou've already chosen the best response.0
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!!
 one year ago

wioBest ResponseYou've already chosen the best response.1
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
 one year ago

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

jotopia34Best ResponseYou've already chosen the best response.0
is the answer tan^1e^x\[\tan^1e^x\]
 one year ago

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

jotopia34Best ResponseYou've already chosen the best response.0
yes you are right but is that the antiderivative part?
 one year ago

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

HoaBest ResponseYou've already chosen the best response.0
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
 one year ago

HoaBest ResponseYou've already chosen the best response.0
oh I'm sorry, arctan (e)  arctan 1. not as above. terribly sorry
 one year ago
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