∫dx/[x√(2x-1)]
let
x= (1/2) (secy)^2
dx = (secy)^2.(tany) dy
∫dx/[x√(2x-1)]
=2∫ dy
=2y + C
=2arccos (1/√(2x)) + C专科生理解不了专科生理解不了。答案是2arctan√(2x-1)+C∫dx/[x√(2x-1)]
let

x= (1/2) (secy)^2
dx = (secy)^2. (tany) dy

∫dx/[x√(2x-1)]

=∫ (secy)^2. (tany) dy / [(1/2) (secy)^2 . √ (secy^2 -1) ]
=∫ (secy)^2. (tany) dy / [(1/2) (secy)^2 . tany ]
=2∫ dy
=2y + C
=2arccos (1/√(2x)) + C
=2arctan√(2x-1) + C