设1/(x^3+1)=a/(x+1)+(bx+c)/(x^2-x+1)
则a+b=0 b+c-a=0 a+c=1
a=1/3 b=-1/3 c=2/3
1/(x^3+1)=1/3(x+1)-(2x-1)/6(x^2-x+1)
+1/2[(x-1/2)^2+3/4]
∴∫dx/(1+x^3)=(1/6)In[(x+1)^2/(x^2-x+1)]
+[1/3^(1/2)]arctan[(2x-1)/3^(1/2)]+C
1/(x^3+1)求不定积分∫(1/x^3+1)d(x)
设1/(x^3+1)=a/(x+1)+(bx+c)/(x^2-x+1)
则a+b=0 b+c-a=0 a+c=1
a=1/3 b=-1/3 c=2/3
1/(x^3+1)=1/3(x+1)-(2x-1)/6(x^2-x+1)
+1/2[(x-1/2)^2+3/4]
∴∫dx/(1+x^3)=(1/6)In[(x+1)^2/(x^2-x+1)]
+[1/3^(1/2)]arctan[(2x-1)/3^(1/2)]+C