第1个回答 2021-12-06
(1)
let
u=x^(1/6)
du = (1/6)x^(-5/6) dx
dx = 6u^5 du
∫ dx/[√x.(1+ x^(1/3)) ]
=∫ 6u^5 du/[ u^3.(1+ u^2) ]
=6∫ u^2 /(1+ u^2) du
=6∫ [1 - 1/(1+ u^2)] du
=6(u -arctanu ) + C
=6{ x^(1/6) -arctan[x^(1/6)] } + C
(2)
∫(0->1) xe^x dx
=∫(0->1) x de^x
=[xe^x]|(0->1) - ∫(0->1) e^x dx
=e - [e^x]|(0->1)
=e- (e-1)
=1