利用分部积分法求解不定积分：∫dx/（x³+1），急求详细解答过程！！！详解！

Answer 1

因为
x^3+1=(x+1)(x²-x+1)
所以
1/(x^3+1)=A/(x+1)+(Bx+C)/(x²-x+1)
A(x²-x+1)+(Bx+C)(x+1)=1
(A+B)x²+(B+C-A)x+A+C=1
A+B=0
B+C=A
A+C=1
A=1/3
B= -1/3
C=2/3
所以
∫1/(x^3+1)dx
=∫［1/3*1/(x+1)+(-1/3*x+2/3)/(x²-x+1)］dx
=1/3*ln|x+1|-1/6*∫(2x-4)/(x²-x+1)dx
=1/3*ln|x+1|-1/6*∫(2x-1)/(x²-x+1)dx+1/2*∫1/(x²-x+1)dx
=1/3*ln|x+1|-1/6*ln(x²-x+1)+1/2*∫1/[(x-1/2)²+(√3/2)²]d(x-1/2)
=1/3*ln|x+1|-1/6*ln(x²-x+1)+1/2*√3/2*arctan[(x-1)/(√3/2)]+C
=1/3*ln|x+1|-1/6*ln(x²-x+1)+1/√3*arctan(2x-1)/√3+C

追答

其中
∫1/(x²+a²)dx
(x=atanu)
=∫1/a²sec²u*asec²udu
=1/a*u+C
=1/a*arctanx/a+C

ln|x²-x+1|绝对值可以直接去掉，x²-x+1=(x-1/2)²+3/4＞0

那里写错了是2/√3

1/2*2/√3*arctan[(x-1)/(√3/2)]

Answer 2

对分母分解因式 然后拆开