换元积分法
令√(x-3)=t x=t^2+3 dx=2tdt
原式=∫(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