求不定积分∫lnx/x√1+lnx dx

∫lnx/[x√(1+lnx)] dx
令 t = √(1+lnx) ,则 lnx = t^2 - 1 ,
x = e^(t^2 - 1) ,代入得
∫lnx/[x√(1+lnx)] dx
= ∫lnx/[√(1+lnx)] d(lnx)
=∫(t^2 - 1)/t ·d(t^2 - 1)
=2∫(t^2 - 1) dt
=(2t^3)/3 - 2t + C
=2/3·[√(1+lnx)]^3 - 2√(1+lnx) + C