如何求函数的极值？

Question

如题所述

Answer 1

∫(1-x)/(1+x^3)dx
这个就需要用因式分解
1+x^3=(1+x)(x^2-x+1)
将（1-x）化成这两个因式的加和
（1-x）=（2/3)(x^2-x+1)-(1/3)(2x-1)(x+1)
∫(1-x)/(1+x^3)dx
=∫[(2/3)(x^2-x+1)-(1/3)(2x-1)(x+1)]/(1+x^3)
dx
=(2/3)∫1/(x+1)dx
-
(1/3)
∫[(2x^2-2x+2)+(3x-3)]/(x^2-x+1)
dx
=(2/3)
ln(x+1)-(2/3)x+(1/2)∫1/(x^2-x+1)d(x^2-x+1)+
(√3/3)arctan[(2x-1)/√3]
=(2/3)
lni
x+1i-(2/3)x+(1/2)lnix^2-x+1i+(√3/3)arctan[(2x-1)/√3]+c
解答完毕，请指教，真麻烦啊呀