计算定积分:∫(0,9)dx/1+√x
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计算定积分:∫(0,9)dx/1+√x
![计算定积分:∫(0,9)dx/1+√x](/uploads/image/z/278956-28-6.jpg?t=%E8%AE%A1%E7%AE%97%E5%AE%9A%E7%A7%AF%E5%88%86%EF%BC%9A%E2%88%AB%EF%BC%880%2C9%EF%BC%89dx%2F1%2B%E2%88%9Ax)
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
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