这一题，怎么求不定积分，麻烦写下过程

(47)令t=√(x-3),x=t^2+3,dx=2tdt ∫x√(x-3) dx =∫(t^2+3)*t*2tdt =∫(2t^4+6t^2)dt =2/5*t^5+2t^3+C =2/5*(x-3)^(5/2)+2*(x-3)^(3/2)+C