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

[latex]\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=I\\\\x=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}{3\sqrt[3]4}\\\\x^2\, |\; 0=A+B\; ,\; \; B=-A=\frac{1}{3\sqrt[3]4}\\\\x^0\, |\; 0=\sqrt[3]{16}\cdot A+\sqrt[3]4\cdot C\; ,\; \; C=-\frac{\sqrt[3]{16}A}{\sqrt[3]4}=\frac{1}{3}[/latex] [latex]I=-\frac{1}{3\sqrt[3]4}\int \frac{dx}{x+\sqrt[3]4}+\int \frac{\frac{1}{3\sqrt[3]4}x+\frac{1}{3}}{x^2-\sqrt[3]4x+\sqrt[3]{16}}dx=\\\\=-\frac{1}{3\sqrt[3]4}\cdot ln|x+\sqrt[3]4|+\frac{1}{3\sqrt[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}{3\sqrt[3]4}\cdot ln|x+\sqrt[3]4|+\frac{1}{3\sqrt[3]4}\int \frac{x+\sqrt[3]4}{(x-\frac{1}{\sqrt[3]2})^2+\frac{3}{\sqrt[3]4}}dx=[/latex] [latex]=[\, t=x-\frac{1}{\sqrt[3]2}\, ]=[/latex] [latex]=-\frac{1}{3\sqrt[3]4}\cdot ln|x+\sqrt[3]4|+\frac{1}{3\sqrt[3]4}\int \frac{t+\frac{1}{\sqrt[3]2}+\sqrt[3]4}{t^2+\frac{3}{\sqrt[3]4}}dt=\\\\=-\frac{1}{3\sqrt[3]4}\cdot ln|x+\sqrt[3]4|+\frac{1}{3\sqrt[3]4}\int \frac{t\, dt}{t^2+\frac{3}{\sqrt[3]4}}+\\\\+\frac{1}{3\sqrt[3]4}\int \frac{\frac{8}{\sqrt[3]2}}{t^2+\frac{3}{\sqrt[3]4}}dt=-\frac{1}{3\sqrt[3]4}\cdot ln|x+\sqrt[3]4|+\frac{1}{6\sqrt[3]4}\cdot ln|t^2+\frac{3}{\sqrt[3]4}|+\\\\+\frac{4}{3}\cdot \frac{\sqrt[3]2}{\sqrt3}arctg\frac{\sqrt[3]2\cdot t}{\sqrt3}+C\; ,\; \; t=x-\frac{1}{\sqrt[3]2}[/latex]