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计算定积分:∫(0,9)dx/1+√x
人气:203 ℃ 时间:2020-03-25 04:10:34
解答
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
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