求助求助一道反常积分的计算题

let
u=√(x-2)
2u du = dx
x=2, u=0
x=+∞, u=+∞
∫(2->+∞) dx/[(x+7)√(x-2)]
=∫(0->+∞) 2u du/[(u^2+9).u]
=2∫(0->+∞) du/(u^2+9)
=(2/9)∫(0->+∞) du/[1+(u/3)^2]
=(2/3)[arctan(u/3)]|(0->+∞)
=(2/3)(π/2)
=π/3