# What is the integral of ##(e^(2x))/(1+e^(2x))##?

##f(x)={e^{2x}}/{1+e^{2x}}##

##I(x) = int{e^{2x} dx}/{1+e^{2x}}##

##u(x) = 2x -> dx = 1/2du## ##I(u)=1/2 int{e^u du}/{1+e^u}##

##v(u) = e^u -> dv = e^u du## ##I(v) = 1/2 int{dv}/{1+v}##

##w(v) = 1+v -> dw = dv## ##I(w) = 1/2 int{1}/{w}dw = 1/2 ln(w)+c, ” ” w>0##

Substituting back in: ##I(v) = 1/2 ln(1+v)+c, ” ” v> -1## ##I(u) = 1/2 ln(1+e^u), ” ” u in R##  ##I(x) = 1/2 ln(1+e^{2x})##

I cant think of a simplification of this function. Note that is is positive-definite, has no extrema, for large positive x, ##I(x) approx x## and ##lim_{x->-infty}I(x)=1/2 ln{2} approx 0.692##

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 ##e^x## is positive for all real numbers. You could hypothetically extend this into complex numbers but there are all sorts of complications involved when applying this question to complex variables  ‘large’ here isnt all that large really – because ##e^{2x}## moves extremely quickly, both limits become good approximations by about ##|x| = 2##  at ##x=2##, ##I(x) = 2.0091##, giving the approximation an accuracy of 99.5% at ##x=-2##, ##I(x) = 0.0091##