Question

Integral from 1 to e of
(e^2x)/(1+e^x)

Answer 1

you would solve this using u-substition

Answer 2

but its a little confusing...with the e^2x since if you set u=1+e^x, and du=e^xdx and that wont cancel out e^2x

Answer 3

actually the problem is a improper integral from negative infinity to infinity sorry

Answer 4

so you have to remember the properties of powers...which e^2x=e^2*e^x and e^2 is a number so you can take it out to find the integral

Answer 5

can you get it from here?

Answer 6

so the u would be (1+e^x)

Answer 7

yup

Answer 8

thanks

Answer 9

e^2x=e^2*e^x is wrong. e^2x = (e^x)^2

Answer 10

if you meant integral (e^[2x])/(1+e^x), then let e^x = u:
(e^[2x])/(1+e^x) becomes (u)/(1+u),
we then do a silly algebra trick,
(u)/(1+u) = (u+[1-1])/(1+u)
which leads to
integral ( 1 - 1/(1+u) ) = u - integral ( 1/(1+u) )
= u - ln(1+u)
substituting e^x = u:
e^x - ln(1+e^x)

Answer 11

perhaps on a silly note it also takes the following form:
\[ \sqrt{(e^{t/2})}* e^{\cosh ^{-1}\sqrt{(e ^{t/2}})}\]

Answer 12

oops, t not t/2

Answer 13

let e^x=u du=e^xdx therefore dx=du/e^x but e^x=u thus dx=du/u proceed....
sol'n e-e^e+ln((1-e)/(1-e^e))