Question

1加x三次方倒数原函数

Answer 1

先分解部分分式：

1/(1+x^3)=1/[(1+x)(1-x+x^2)]=a/(1+x)+(bx+c)/(1-x+x^2)

去分母：1=a(1-x+x^2)+(bx+c)(1+x)

令x=-1,得：1=3a, 得a=1/3

令x=0,得：1=a+c, 得：c=2/3

令x=1,得: 1=a+2(b+2/3),得：b=-1/3

故1/(1+x^3)=1/3[1/(x+1)-(x-2)/(1-x+x^2)]

而(x-2)/(1-x+x^2)=(x-1/2-3/2)/[(x-1/2)^2+3/4]=(x-1/2)/[(x-1/2)^2+3/4]-(3/2)/[(x-1/2)^2+3/4]

积分即得原函数=1/3ln|x+1|-1/3[1/2ln(1-x+x^2)-√3arctan((2x-1)/√3)]+C