Question

∫根号(x-1)lnxdx用换元法和分部积分法求

Answer 1

∫ √(x-1) .lnx dx
=(2/3)∫ lnx d(x-1)^(3/2)
=(2/3)(x-1)^(3/2). lnx -(2/3)∫ [(x-1)^(3/2) /x ] dx
=(2/3)(x-1)^(3/2). lnx -(2/3)∫ [ √(x-1) - √(x-1) /x ] dx
=(2/3)(x-1)^(3/2). lnx - (x-1)^(3/2) + (3/2)∫[ √(x-1) /x ] dx
let
x= (secy)^2
dx = 2tany (secy)^2 dy

∫[ √(x-1) /x ] dx
=2∫ (tany)^2 dy
=2( tany - y ) + C'
=2 {√(x-1) - arctan[√(x-1)] } + C'
∫ √(x-1) .lnx dx
=(2/3)(x-1)^(3/2). lnx - (x-1)^(3/2) + (3/2)∫[ √(x-1) /x ] dx
=(2/3)(x-1)^(3/2). lnx - (x-1)^(3/2) + 3{√(x-1) - arctan[√(x-1)] } + C