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OpenStudy (anonymous):

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

OpenStudy (anonymous):

you would solve this using u-substition

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

can you get it from here?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yup

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

oops, t not t/2

OpenStudy (anonymous):

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))

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