((根号下x-1)×arctan根号下x-1)/x的不定积分如何计算

u=√(x-1)
2u du = dx
∫ √(x-1) . arctan√(x-1) /x dx
= ∫ [u . arctanu /(u^2 +1) ] ( 2u du)
= 2∫ [u^2 . arctanu /(u^2 +1) ]du
= 2∫ [1 - arctanu/(u^2 +1) ]du
=2[ u - (1/2)(arctanu)^2 ] + C
=2u - (arctanu)^2 + C
=2√(x-1) - [ arctan√(x-1) ]^2 + C本回答被网友采纳