不定积分：∫ 1⼀(1-x^3) dx 有什么好方法

-1/(x-1)(x²+x+1)
设
=A/(x-1)+(Bx+C)/(x²+x+1)
通分后计算分母得1，所以
A(x²+x+1)+(Bx+C)(x-1)=1
(A+B)x²+(A-B+C)x+A-C=1

A+B=0
A-B+C=0
A-C=1
解得A=1/3,B=-1/3,C=-2/3
原式=A/(x-1)+(Bx+C)/(x²+x+1)
=[1/(x-1)-(x+2)/(x²+x+1)]/3

∫ [1/(x-1)-(x+0.5+1.5)/(x²+x+1)]/3 dx
=∫ [1/(x-1)-(x+0.5)/(x²+x+1)-1.5/(x²+x+1)]/3 dx
=(1/3)[ln|x-1|-0.5ln(x²+x+1)] - ∫0.5/[(x+1/2)²+3/4]dx

重点解决
∫0.5/[(x+1/2)²+3/4]dx
设(x+1/2)=[(根号3)/2]tant
dx=[(根号3)/2]sec²t dt

∫0.5/[(x+1/2)²+3/4]dx
=0.5 ∫[1/(3/4sec²t)][(根号3)/2]sec²t dt
=0.5*4*(根号3)/(3*2)∫1 dt
=(根号3/3)t+C

tant=(2x+1)/(根号3)
t=arctan[(2x+1)/(根号3)]

∫0.5/[(x+1/2)²+3/4]dx
=((根号3)/3)*arctan[(2x+1)/(根号3)] +C

带回
(1/3)[ln|x-1|-0.5ln(x²+x+1)] - ∫0.5/[(x+1/2)²+3/4]dx
=(1/3)[ln|x-1|-0.5ln(x²+x+1)] - ((根号3)/3)*arctan[(2x+1)/(根号3)] +C

没什么好办法,对1/(1-x^3)分解得:
1/(1-x^3)=1/3(1-x)+(x+2)/3(1+x+x^2)
所以原式=∫1/3(1-x)dx+∫(x+2)/3(1+x+x^2)dx
=-ln | x-1 |/3+ln(x^2+x+1)/6+2arctan[(2x+1)/根号3]/根号3+C