Найти неопределенный интеграл
∫xdx/(x^3+4)

$int frac{x, dx}{x^3+4}=int (frac{A}{x+sqrt[3]4}+frac{Bx+C}{x^2-sqrt[3]4x+sqrt[3]{16}})dx=Ix=A(x^2-sqrt[3]4x+sqrt[3]{16})+(Bx+C)(x+sqrt[3]4)x=-sqrt[3]4; to ; A=frac{x}{x^2-sqrt[3]4x+sqrt[3]{16}}=frac{-sqrt[3]4}{sqrt[3]{16}+sqrt[3]{16}+sqrt[3]{16}}=-frac{1}{3sqrt[3]4}x^2, |; 0=A+B; ,; ; B=-A=frac{1}{3sqrt[3]4}x^0, |; 0=sqrt[3]{16}cdot A+sqrt[3]4cdot C; ,; ; C=-frac{sqrt[3]{16}A}{sqrt[3]4}=frac{1}{3}$

$I=-frac{1}{3sqrt[3]4}int frac{dx}{x+sqrt[3]4}+int frac{frac{1}{3sqrt[3]4}x+frac{1}{3}}{x^2-sqrt[3]4x+sqrt[3]{16}}dx==-frac{1}{3sqrt[3]4}cdot ln|x+sqrt[3]4|+frac{1}{3sqrt[3]4}int frac{x+sqrt[3]4}{(x-frac{sqrt[3]4}{2})^2-frac{sqrt[3]{16}}{4}+sqrt[3]{16}}dx==-frac{1}{3sqrt[3]4}cdot ln|x+sqrt[3]4|+frac{1}{3sqrt[3]4}int frac{x+sqrt[3]4}{(x-frac{1}{sqrt[3]2})^2+frac{3}{sqrt[3]4}}dx=$

$=[, t=x-frac{1}{sqrt[3]2}, ]=$

$=-frac{1}{3sqrt[3]4}cdot ln|x+sqrt[3]4|+frac{1}{3sqrt[3]4}int frac{t+frac{1}{sqrt[3]2}+sqrt[3]4}{t^2+frac{3}{sqrt[3]4}}dt==-frac{1}{3sqrt[3]4}cdot ln|x+sqrt[3]4|+frac{1}{3sqrt[3]4}int frac{t, dt}{t^2+frac{3}{sqrt[3]4}}++frac{1}{3sqrt[3]4}int frac{frac{8}{sqrt[3]2}}{t^2+frac{3}{sqrt[3]4}}dt=-frac{1}{3sqrt[3]4}cdot ln|x+sqrt[3]4|+frac{1}{6sqrt[3]4}cdot ln|t^2+frac{3}{sqrt[3]4}|++frac{4}{3}cdot frac{sqrt[3]2}{sqrt3}arctgfrac{sqrt[3]2cdot t}{sqrt3}+C; ,; ; t=x-frac{1}{sqrt[3]2}$