Question

∫ dx/[(根号内2x-3)+1],用换元法

Answer 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
自己反代吧