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∫arctan√x dx
人气:481 ℃ 时间:2020-02-05 13:33:54
解答
∫arctan√x dx令√x=t,x=t^2,dx=dt^2所以原式=∫arctantdt^2=t^2*arctant-∫t^2/(1+t^2)dt=t^2*arctant-∫(t^2+1-1)/(1+t^2)dt=t^2*arctant-t+arctant+c=xarctan√x-√x+arctan√x+c
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