Question

Evaluate the integral..

Answer 1

is arctan possible here

Answer 2

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

Answer 3

Yes, it is possible

Answer 4

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

Answer 5

it is in fact arctan in one step right?

Answer 6

The problem is I don't know how first time seeing such problem\[\int\limits \frac{du}{a^2+u^2}=\frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right)+C \] took this from my formula ntbk

Answer 7

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

Answer 8

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

Answer 9

@freckles that is clever, i would have taking
\[\lim_{t\to -\infty}\tan^{-1}(e^t)\]

Answer 10

*taken

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

\[\int\limits_{-\infty}^{\infty} \frac{1}{e^{x}+e^{-x} } dx \\ \text{ multiply \top and bottom by } e^{x} \\ \int\limits_{-\infty}^{\infty} \frac{e^{x}}{e^{2x}+1} dx \\ \text{ \let } u=e^{x} \\ du=e^{x} dx \\ \\ \int\limits_0^\infty \frac{du }{u^2+1}\]

Answer 15

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

Answer 16

Wait I'll analyze it

Answer 17

Why do you have to multiply e^x? Is that algebra or what, you do that if you want to bring something up to numerator?