Question

∫(x^3-3x^2+4x-9)/(x^2+3)dx

Answer 1

(x^3-3x^2+4x-9)/(x^2+3)=x-3+x/(x^2+3)
原式=∫xdx-∫3dx+∫x/(x^2+3) dx
=x^2/2-3x+1/2∫d(x^2+3)/(x^2+3)
=x^2/2-3x+ln(x^2+3)/2+C

Answer 2

∫(x^3-3x^2+4x-9)/(x^2+3)dx
= ∫(x-3) dx +∫ x/(x^2+3) dx
= x^2/2 - 3x +∫ x/(x^2+3) dx
let
x = √3 tany
dx =√3 (secy)^2dy
∫ x/(x^2+3) dx
=∫ [√3 tany / (3secy)] √3 (secy)^2dy
= ∫tanysecydy
=secy
= (x^2/3 - 1)^(1/2)
∫(x^3-3x^2+4x-9)/(x^2+3)dx
=x^2/2 - 3x +∫ x/(x^2+3) dx
=x^2/2 - 3x+(x^2/3 - 1)^(1/2)+C

Answer 3

∫(x^3-3x^2+4x-9)/(x^2+3)dx=∫(x^3-3x^2+3x-9)/(x^2+3)dx+∫x/(x^2+3)dx
=∫(x^2+3)(x-3)/(x^2+3)dx+∫x/(x^2+3)dx
=∫(x-3)dx+∫x/(x^2+3)dx
=1/2x^2-3x+1/2ln(x^2+3) +C