Question

求不定积分∫［3/(x^3+1)］dx

Answer 1

3/(x^3+1)=a/(x+1)+(bx+c)/(x^2-x+1)
两边同乘以x+1,再令x=-1
1=a
令x=0,得
3=1+c
c=2
令x=1
3/2=1/2 +(b+c)/1
b+c=1
b=-1
所以
3/(x^3+1)=1/(x+1)+(-x+2)/(x^2-x+1)
从而
原式=∫1/(x+1)dx-∫(x-2)/(x^2-x+1)dx
=ln|x+1|-∫(x-1/2 -3/2)/(x^2-x+1)dx
=ln|x+1|-1/2ln(x^2-x+1)+3/2∫1/(x^2-x+1) dx
=ln|x+1|-1/2ln(x^2-x+1)+3/2∫1/[(x-1/2)^2+3/4] dx
=ln|x+1|-1/2ln(x^2-x+1)+3/2×1/(√3/2)arctan(x-1/2)/(√3/2)+c
=ln|x+1|-1/2ln(x^2-x+1)+√3arctan(2x-1)/(√3)+c