求这道题的不定积分，需要详细过程，谢谢啦

Question

如题所述

Answer 1

∫dx/(√(2x-3)+1)
令√(2x-3)=t
x=(t^2+3)/2
dx=tdt
∫ dx/[√(2x-3)+1]
=∫t/(t+1)dt
=∫[1-1/(t+1)]dt
=t-ln(t+1)+C
=√(2x-3)-ln(√(2x-3)+1)+C