∫(x^3+1)/(x(1-x^3))dx

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∫(x^3+1)/(x(1-x^3))dx∫(x^3+1)/(x(1-x^3))dx∫(x^3+1)/(x(1-x^3))dx(1+x³)/[x(1-x³)]=(1+x³

∫(x^3+1)/(x(1-x^3))dx
∫(x^3+1)/(x(1-x^3))dx

∫(x^3+1)/(x(1-x^3))dx
(1+x³)/[x(1-x³)]
=(1+x³)/[x(1-x)(1+x+x²)]
令(1+x³)/[x(1-x)(1+x+x²)]=A/(1+x+x²)+B/(1-x)+C/x
,应用待定系数法,解得A=(1/3)(-4x-2),B=2/3,C=1
原式变为(-2/3)∫(2x+1)/(x²+x+1) dx-(2/3)∫dx/(x-1)+∫dx/x
=(-2/3)∫d(x²+x+1)/(x²+x+1)-(2/3)∫dx/(x-1)+∫dx/x
=(-2/3)ln|x²+x+1|-(2/3)∫ln|x-1|+ln|x|
=lnx-(2/3)ln|x³-1|+C

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