Question

求不定积分∫[1/(1+x^3)]dx 要步骤

Answer 1

拆成两部分，再积分

详情如图所示

Answer 2

1+x^3=(x+1)(x^2-x+1)
用待定系数法：A/(x+1)+(Bx+c)/(x^2-x+1)=1/(x+1)(x^2-x+1)
得A=1/3,B=-1/3,C=2/3
所以∫[1/(1+x^3)]dx
=1/3∫(1/(x+1))dx-1/3∫((x-2)/(x^2-x+1))dx
其中1/3∫(1/(x+1))dx=1/3ln|x+1|+c
因为d（x^2-x+1）=(2x-1)dx,所以x-2=1/2(2x-1)-3/2
∫((x-2)/(x^2-x+1))dx=1/2∫(d(x^2-x+1)/(x^2-x+1))-3/2∫(1/(x^2-x+1))dx
其中∫(d(x^2-x+1)/(x^2-x+1))=ln|x^2-x+1|+c
∫(1/(x^2-x+1))dx=∫(dx/((x-1/2)^2+(根号3/2)^2))
因为∫（dx/(x^2+a^2)）=(1/a)arctan(x/a)
所以∫(1/(x^2-x+1))dx=∫(dx/((x-1/2)^2+(根号3/2)^2))
=（2/根号3）arctan（(x-1/2)/(根号3/2)）+c
在乘上系数，整理∫[1/(1+x^3)]dx=1/3ln|x+1|-1/6|x^2-x+1|+(1/根号3)arctan((2x-1)/根号3)+c

Answer 3

∫(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