Question

这题不会了，求解

Answer 1

f(x) = ∫(1->x) lnt/(1+t) dt

To prove: f(x) + f(1/x) =(1/2)(lnx)^2
solution:
let
u=1/t
du = -(1/t^2) dt
dt = -du/u^2
t=1, u=1
t=1/x , u= x
f(1/x)
=∫(1->1/x) lnt/(1+t) dt
=∫(1->x) [ln(1/u)/(1+1/u)] .(-du/u^2)
=∫(1->x) lnu/[u(1+u)] du
=∫(1->x) lnt/[t(1+t)] du
f(x)+f(1/x)
=∫(1->x) lnt/(1+t) dt +∫(1->x) lnt/[t(1+t)] du
=∫(1->x) (t+1).lnt/[t(1+t)] . dt
=∫(1->x) lnt/t dt
=∫(1->x) lnt dlnt
=(1/2)[ (lnt)^2]|(1->x)
=(1/2)(lnx)^2