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大神帮忙看看这道定积分的题谢谢
3个回答
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用换元法
设√x=t(t≥o),x=t^2,dx=2tdt,则
原式=∫t/(1+t)2tdt=2∫t^2/(1+t)dt=2∫(t^2-1+1)/(t+1)dt=2∫(t-1)+2/(t+1)dt
=t^2-2t+2ln(t+1)
所以原式=x-2√x+2丨n(√x+1)
=2-2√2+2|n(√2+1)-(1-2+2|n2)
=3-2√2+|n(√2+1)^2/4
设√x=t(t≥o),x=t^2,dx=2tdt,则
原式=∫t/(1+t)2tdt=2∫t^2/(1+t)dt=2∫(t^2-1+1)/(t+1)dt=2∫(t-1)+2/(t+1)dt
=t^2-2t+2ln(t+1)
所以原式=x-2√x+2丨n(√x+1)
=2-2√2+2|n(√2+1)-(1-2+2|n2)
=3-2√2+|n(√2+1)^2/4
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