搜搜考试网设f(x)的原函数F(x)>0,且f(x)F(x)=1/(e^x+e^-x),F(0)=√(π/2),求证f(x)=e^
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设f(x)的原函数F(x)>0,且f(x)F(x)=1/(e^x+e^-x),F(0)=√(π/2),求证f(x)=e^x/(1+e^2x)√(2arctane^x)
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f(x)F(x)=1/(e^x+e^-x)
∫ f(x)F(x) dx = ∫ dx/[e^x+e^(-x)]
let
y = e^x
dy = e^xdx
∫ dx/[e^x+e^(-x)]
= ∫dy/(y^2+1)
= arctany + C1
= arctan(e^x) + C1
∫ f(x)F(x) dx = ∫ dx/[e^x+e^(-x)]
[F(x)]^2/2 =arctan(e^x) + C1
put x=0
π/4 = π/4 + C1
C1=0
[F(x)]^2/2 =arctan(e^x)
F(x) = √[ 2arctan(e^x)]
f(x) =F'(x)
=(1/[2√[ 2arctan(e^x)] ] ) .[2e^x/(1+e^(2x)]
= e^x/[(1+e^2x)√(2arctan(e^x))]
∫ f(x)F(x) dx = ∫ dx/[e^x+e^(-x)]
let
y = e^x
dy = e^xdx
∫ dx/[e^x+e^(-x)]
= ∫dy/(y^2+1)
= arctany + C1
= arctan(e^x) + C1
∫ f(x)F(x) dx = ∫ dx/[e^x+e^(-x)]
[F(x)]^2/2 =arctan(e^x) + C1
put x=0
π/4 = π/4 + C1
C1=0
[F(x)]^2/2 =arctan(e^x)
F(x) = √[ 2arctan(e^x)]
f(x) =F'(x)
=(1/[2√[ 2arctan(e^x)] ] ) .[2e^x/(1+e^(2x)]
= e^x/[(1+e^2x)√(2arctan(e^x))]
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