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|dw:1359688972018:dw|

hmm

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

sorry I made a mistake its e^2x

oh then

Try u sub

\(u=e^x+1\)

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

Okay, what exactly is the equation?

|dw:1359689045480:dw|

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

What is the equation?

I do not know what the correct equation is.

the euation is above

There are MANY equations above.

Which one is correct?

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

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

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

yes

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

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

and then satellite makes it look ez <.<

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

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

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

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

oh it is a definite integral

This is scandalous!!

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

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

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

yes it certainly is

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

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

laws 'o exponents

it amazes me that no one ever pointed that out before

lol. Laws 'o fuh shizzle

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

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

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

You're confusing multiplication with exponents

jotopia look at this my pic below

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

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

|dw:1359689988230:dw|

oh yeah!! I get it now

that is key. You are all fantabulous

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

so, no u sub?

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

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

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

yes you are right but is that the antiderivative part?

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

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