Question

大学数学不定积分问题

Answer 1

∫ [ √x/(1-x^(1/3) ] dx
=∫ [-x^(1/6) + x^(1/6)/(1-x^(1/3)] dx
= -(6/7)x^(7/6) +∫[ x^(1/6)/(1-x^(1/3)] dx

let
x^(1/6) = siny
(1/6)x^(-5/6) dx = cosy dy
dx = 6(siny)^5. cosy dy

∫[ x^(1/6)/(1-x^(1/3)] dx
=6∫[ (siny)^6/cosy] dy
=6∫ [( 1- (cosy)^2 )^3/cosy ] dy
=6∫ [(1/cosy) -3 +3(cosy) -(cosy)^2 ] dy
=3∫ [ (2/cosy) -6 +6(cosy) -1-cos2y ] dy
= 3[ 2ln|secy + tany| - 6y +6siny - y - sin2y/2 ] + C'
= 3[ 2ln|x^(-1/6)+ √(1-x^(1/3))| - 6arcsin(x^(1/6)) +6x^(1/6) - x^(1/6).√(1-x^(1/3)) ] +C'

∫ [ √x/(1-x^(1/3) ] dx
=-(6/7)x^(7/6) +∫[ x^(1/6)/(1-x^(1/3)] dx
=-(6/7)x^(7/6)
+3[ 2ln|x^(-1/6)+ √(1-x^(1/3))| - 6arcsin(x^(1/6)) +6x^(1/6) - x^(1/6).√(1-x^(1/3)) ] +C

Answer 2

简单

追答

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