设 u=arctan [x^(1/2)],
则du =(1/2)x^(-1/2)/ (1+x)
∫arctan [x^(1/2)] / [ (1+x) x^(1/2) ]dx = ∫ 2udu =u^2 + C
= { arctan [x^(1/2)] }^2 + C
求不定积分arctan(x^1/2)/((x^1/2)×(1+x))dx
求不定积分arctan(x^1/2)/((x^1/2)×(1+x))dx
数学人气：545 ℃时间：2020-02-02 22:10:25
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