arctan(x)^1/2的不定积分
人气:243 ℃ 时间:2020-06-18 16:44:39
解答
√x=u,dx=2udu
=∫2arctanudu^2
=2u^2arctanu-2∫u^2/(1+u^2)*du
=2u^2arctanu-2∫(u^2+1-1)/(1+u^2)*du
=2u^2arctanu-2u+2arctanu+C
=2xarctan√x-2√x+2arctan√x+C
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