被积函数是1/(1+x^3)的不定积分怎么解

∫1/（1+x^3）dx=∫1/(x+1)*(x^2-x+1)dx=1/3∫1/(x+1)dx-1/3∫(x-2)dx/(x^2-x+1)=1/3∫1/(x+1)dx-1/6∫d(x^2-x+1)/(x^2-x+1)+1/2∫dx/[(x-1/2)^2+3/4]=1/3∫1/(x+1)dx-1/6∫d(x^2-x+1)/(x^2-x+1)+1/√3∫d(2/√3*x-1/√3)/[(2/√3*x-1/√3)^2+1]=1/6ln[(x+1)^2/x^2-x+1]+1√3/arctan[2/√3*x-1/√3)^2+1]+c 再问： 变换为1/3∫1/(x+1)dx-1/3∫(x-2)dx/(x^2-x+1)这一步是怎么想到的呀？