∫1/(1+√x) dx 用第二类换元法求不定积分过程,
人气:489 ℃ 时间:2020-01-29 22:51:22
解答
令√x=t,则x=t^2
dx=d(t^2)=2tdt
∴原式=∫1/(1+√x) dx
=∫2t/(1+t) dt
=∫(2(t+1)-2)/(1+t)dt
=∫2dt-∫2/(1+t)dt
=2t-2ln|t+1|+C
=2√x-2ln|√x+1|+C
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