求不定积分∫[(arctan√x)/(√x(1+x))]dx

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求不定积分∫[(arctan√x)/(√x(1+x))]dx
求不定积分∫[(arctan√x)/(√x(1+x))]dx

求不定积分∫[(arctan√x)/(√x(1+x))]dx
令y=√x,代入上式得:
2∫y[(arctany)/(y(1+y^2))]dy
=2∫[(arctany)/(1+y^2)]dy
=2∫(arctany)d(arctany)=(arctany)^2+c
=(arctan√x)^2+c