答案是(1/3)ln(x-1)-(1/6)ln(x²+x+1)-(2/√3)arctan[(2x+1)/√3]+C

過程:

∫1/(x³-1)

dx

=∫1/(x-1)(x²+x+1)

dx

=∫1/3(x-1)-(x+2)/3(x²+x+1)]

dx,總之在/後面的都是分母

=1/3*∫1/(x-1)

dx-1/3*∫(x+2)/(x²+x+1)

dx

=1/3*∫1/(x-1)

d(x-1)-1/3*∫[(2x+1)/2(x²+x+1)+3/2(x²+x+1)]

dx

=1/3*ln(x-1)-1/6*∫(2x+1)/(x²+x+1)

dx-1/2*∫1/(x²+x+1)

dx

=1/3*ln(x-1)-1/6*∫(2x+1)d(x²+x+1)/(x²+x+1)(2x+1)-1/2*∫1/[(x+1/2)²+3/4]

dx

=1/3*ln(x-1)-1/6*ln(x²+x+1)-1/2*∫1/[(x+1/2)²+3/4]

d(x+1/2)

=1/3*ln(x-1)-1/6*ln(x²+x+1)-√(4/3)*arctan[√(4/3)*(x+1/2)]+C

=1/3*ln(x-1)-1/6*ln(x²+x+1)-2/√3*arctan[(2x+1)/√3)]+C

=(1/3)ln(x-1)-(1/6)ln(x²+x+1)-(2/√3)arctan[(2x+1)/√3]+C