Question

I encountered a question I couldn't do.

Answer 1

it seemed a bit easy at first, but I just....
\[\large \int\limits_{ }^{ } \frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}dx\]

Answer 2

I though of simplifying it to,
\[\large \int\limits_{ }^{ } \frac{e^{x^2}+1}{e^{x^2}-1}dx\]but does it make it any better ?

Answer 3

u=e^x
dx=e^x dx >>> dx=1/ln(u) du

Answer 4

\[\large \int\limits_{ }^{ } \frac{1}{\ln(u)}~\frac{u^2+1}{u^2-1}dx\]

Answer 5

Hmm I think I see a nice easy U-substitution from the start :)

Answer 6

h that shold be du

Answer 7

well, I am kind of going to have to simplify that as I did in my second step either way.
Doesn't matter if it is with u's or with e^x 's....

Answer 8

but if what I did is correct, then it shouldn't be hard with by parts

Answer 9

HI!!

Answer 10

\[\Large\rm \large \int\limits\limits_{ }^{ } \frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}dx=\int\limits \frac{\color{royalblue}{(e^x+e^{-x})dx}}{\color{orangered}{e^x-e^{-x}}}\]

Answer 11

make u the whole denominator, a miracle will occur

Answer 12

ohhh

Answer 13

yes, zepdrix..

Answer 14

how have I not thoguht of this, I am so dump. ty!

Answer 15

Those darn exponentials are so silly >.<
Easy to miss those things!

Answer 16

so the answer is
\[\ln\left| e^x-e^{-x}\right|+C\]

Answer 17

tnx, I was over-thinking it.

Answer 18

yay good job \c:/