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

- jotopia34

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

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- Goten77

|dw:1359688972018:dw|

- Goten77

hmm

- jotopia34

\[\int\limits_{0}^{1}e^x/e ^{2x}+1\]

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## More answers

- jotopia34

sorry I made a mistake its e^2x

- Goten77

oh then

- anonymous

Try u sub

- anonymous

\(u=e^x+1\)

- jotopia34

but it I do u=e^2x, du= 2e^2x and I still can't get rid of the e^x

- anonymous

Okay, what exactly is the equation?

- Goten77

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- jotopia34

nope. I still dont have it.....

- anonymous

What is the equation?

- anonymous

I do not know what the correct equation is.

- jotopia34

the euation is above

- anonymous

There are MANY equations above.

- anonymous

Which one is correct?

- jotopia34

\[\int\limits_{0}^{1}e^x/e ^{2x}+1\]

- anonymous

So it isn't this?\[\Large
\int_0^1 \frac{e^2}{e^{2x}+1} dx
\]

- anonymous

\[
\Large
\int_0^1 \frac{e^x}{e^{2x}+1} dx \]

- jotopia34

yes

- anonymous

think of
\[\int\frac{x}{1+x^2}dx\]

- Goten77

|dw:1359689203929:dw| eh i think im doing something wrong....

- Goten77

and then satellite makes it look ez <.<

- jotopia34

wait, I thought you can't have a u and an x in the equation

- Goten77

thats right jotopia... thats y i knew i did it wrong XD

- anonymous

\[u=e^x,du=e^xdx\] you get
\[\int\frac{du}{1+u^2}\]

- jotopia34

I don't see what sattelite said because how can e^2x be like x^2

- anonymous

oh it is a definite integral

- jotopia34

This is scandalous!!

- anonymous

\(u(0)=1,u(1)=e\) you get
\[\int_1^e\frac{du}{1+u^2}\]

- anonymous

Since \(u=e^{2x}=(e^x)^2 \implies \sqrt{u}=e^x\), @Goten77 's method isn't hopeless.

- anonymous

anti derivative of \(\frac{1}{1+u^2}\) is \(\tan^{-1}(u)\)

- Goten77

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

oh, e^2x is the same as saying e^x squared????/

- anonymous

yes it certainly is

- jotopia34

like how cos^2x is (cosx)^2

- anonymous

\((e^x)^2=e^{2x}\)

- anonymous

laws 'o exponents

- jotopia34

it amazes me that no one ever pointed that out before

- jotopia34

lol. Laws 'o fuh shizzle

- anonymous

and also you are right, \(\cos^2(x)=(\cos(x))^2\)

- anonymous

it is just more common to write \(f^2(x)\) than \((f(x))^2\)

- jotopia34

yes, but its confusing because if you said (e^x)^2, would it be like saying \[e ^{2}x ^{2}\]

- anonymous

You're confusing multiplication with exponents

- Goten77

jotopia look at this my pic below

- jotopia34

hmmm, but isnt (2x)^2 the same as 2^2x^2. And e is a number

- anonymous

\((ex)^2 = e^2x^2\) but \((e^x)^2 =e^{2x}\)

- Goten77

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- jotopia34

oh yeah!! I get it now

- jotopia34

that is key. You are all fantabulous

- anonymous

Anyway, if you sub \(u = e^{2x}+1\) then \(e^x = \sqrt{u^2-1}\) which is also complicated

- jotopia34

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

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!!

- anonymous

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

so, no u sub?

- anonymous

Well you two two substitutions:\[
1)\quad u= e^x \\
2)\quad u=\tan(\theta)
\]

- jotopia34

is the answer tan^-1e^x\[\tan^-1e^x\]

- anonymous

That's the anti-derivative, you have to put in the limits.

- jotopia34

yes you are right but is that the antiderivative part?

- anonymous

Yeah, though I'd like to see your work.

- anonymous

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

- anonymous

oh I'm sorry, arctan (e) - arctan 1. not as above. terribly sorry

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