求不定积分∫x/√(x-3) dx 麻烦写下具体过程,

Question

如题所述

Answer 1

t=√(x-3),x=t^2+3,dx=2tdt
∫x/√(x-3) dx
=∫(t^2+3)/t*2tdt
=∫(2t^2+3)dt
=2/3*t^3+3t+c
=2/3*(x-3)^(3/2)+3*(x-3)^(1/2)+c