大佬们救救我

F(1) =(√2/4)π
f(x).F(x) = arctan√x/[√x.(1+x)]
x=1
f(1).F(1) = (arctan1)/2
f(1).[(√2/4)π] = π/8
f(1)=√2/4
∫f(x).F(x) dx =∫arctan√x/[√x.(1+x)] dx
(1/2)[F(x)]^2
=∫arctan√x/[√x.(1+x)] dx
u=√x
=∫{ arctanu/[u.(1+u^2)] } (2u du)
=2∫arctanu/(1+u^2) du
=[arctanu]^2 + C
=[arctan√x]^2 + C
x=1
(1/2)[F(1)]^2 = π^2/16 + C
(1/2)(1/8)π^2 = π^2/16 + C
C= 0
ie
(1/2)[F(x)]^2 =[arctan√x]^2
[F(x)]^2 =2[arctan√x]^2
[F(x).f(x)]^2 =2[arctan√x]^2 .[f(x)]^2
{arctan√x/[√x.(1+x)]}^2 = 2[arctan√x]^2 .[f(x)]^2
f(x) = 1/[√2.√x.(1+x)]