Question

求不定积分∫dx/(1+x^1/2 )

Answer 1

令√x=t

x=t^2

dx=2tdt

∫dx/(1+x^1/2 )

=∫1/(1+t)*2tdt

=2∫[1-1/(1+t)]dt

=2t-2ln(1+t)+C

自己反代吧