# 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## [1] ##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[2] positive x, ##I(x) approx x## and ##lim_{x->-infty}I(x)=1/2 ln{2} approx 0.692##

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[1] ##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 [2] ‘large’ here isnt all that large really – because ##e^{2x}## moves extremely quickly, both limits become good approximations[3] by about ##|x| = 2## [3] at ##x=2##, ##I(x) = 2.0091##, giving the approximation an accuracy of 99.5% at ##x=-2##, ##I(x) = 0.0091##