计算定积分：∫（0,9）dx/1+√x

题目

计算定积分：∫（0,9）dx/1+√x

答案

令√x=t
x=0,t=0,x=0,t=3
x=t^2,dx=2tdt
∫[0,9]dx/(1+√x)
=∫[0,3] 2tdt/(1+t)
=2∫[0,3] [1-1/(1+t)]dt
=2[t-ln(1+t)] [0,3]
=6-2ln4
=6-4ln2