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is arctan possible here

\[\int\limits _{-\infty }^{\infty }\:\left(\frac{dx}{e^x+e^{-x}}\right)\]

Yes, it is possible

I would make it more obvious that is so by multiplying e^x on top and bottom

it is in fact arctan in one step right?

\[\int\limits_{}^{}\frac{ du}{u^2+1} \text{ where } u=e^x \]

the lower limit of integration changes since as x goes to -inf
then u goes to 0

*taken

what about e^-x do I have to bring that up?

who cares we aren't English snobs we are math snobs @satellite73

\[\frac{e^x}{e^x}\]

@arvnoodle that have written out does it make more sense or still no sensE?

Wait I'll analyze it

|dw:1438745594477:dw|

yes that is algebra

I noticed the lower limit changed to 0 why is that

as x approaches -inf
e^x approaches 0

Oh okay tried it in my calcu it reaches 0 so that make sense

|dw:1438745936722:dw|
notice as x gets infinitely negative large the y values are getting close to 0

\[\int\limits _0^{\infty }\:\frac{du}{u^2+1}\] sooo

\[\frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right)+C\]

is this the one? is my formula right?

but a is 1
and we don't need the C because we have a definite integral

\[\int\limits_0^\infty \frac{1}{u^2+1} du=\lim_{z \rightarrow \infty}[\arctan(u)|_0^{z}]\]

Oh okay so arctan(e^x) ? with limits of 0 and inf

well if you change it back to arctan(e^(x))
then you need to go back to the original limits

So will the final answer be infinity?

no

okay let me get this cllear for me

\[\left[\tan^{-1}\left(\frac{e^x}{1}\right)\right]\]

that's right? ^

if we put it back in terms of x then we have to go back to the original limits

okay -inf to +inf

which is fine

but you will have to break that integral up

the -inf turns to 0

Hmmm break?

oh my this problem kind of hard lol sorry for not getting it quickly

the answer is pi/2

yes

I'm unsatisfied because by my logic the e^x reaches inf it means the answer is inf

it doesn't make sense pi/2 to me but ohwell

as x gets large then e^x gets large
then arctan(getting super large)=?

|dw:1438747311900:dw|
look at the graph of y=arctan(x)
as x gets supper large y approaches pi/2

super!

oh I get by graph!!!

oh my does it mean I need to graph when I'm answering this kind of question, kind of tedious lol

but okay thankyou tho

the graph was more of me trying to convince you that it is pi/2
it is not needed in the answer

I got it now I got now!! by chain rule