Question

3/(x^3+1)的不定积分

Answer 1

(x+1)(x^2-x+1)=x^3+1,
3 / [(x+1)(x^2-x+1)]=1/(x+1) + (2-x) / (x^2-x+1)
∫ 3/(x^3+1)dx = ∫ 1/(x+1) + (2-x) / (x^2-x+1) dx
= ∫ 1/(x+1)dx + 1/2* ∫ (2-2x) / (x^2-x+1) dx
=ln|1+x |+ 1/2 * [ ∫ (1-2x) / (x^2-x+1) dx+ ∫ 1 / (x^2-x+1) dx ]
=ln|1+x |+ 1/2 * [ - ∫ 1/ (x^2-x+1) d(x^2-x+1)+ ∫ 1 / [(x-1/2)^2 +3/4 ]d x ]
=ln|1+x |+ 1/2 * [ - ln |x^2-x+1| +2/√3 * ∫ 1 / [ (2x-1)^2 / 3 +1 ]d (2x-1)/√3 ]
=ln|1+x |+ 1/2 * [ - ln |x^2-x+1| +2/√3 *arctan (2x-1)/√3 ] +c