解
令√x=t
则t²=x,dx=2tdt
∴∫dx/(1+√x)
=∫2tdt/(t+1)
=2∫[(t+1)-1]/(t+1)dt
=2∫1-1/(t+1)dt
=2t-2ln|t+1|+C
=2√x-2ln(√x+1)+C=2��1-1/(t+1)dt �����=2��[1-1/(t+1)]dt����1��ȥ(t+1��֮1)