Помогите решить интеграл Интеграл x/(x^3+8) dx

Решите задачу:

$\displaystyle\int\frac{xdx}{x^3+8}=-\frac{1}{6}\int\frac{dx}{x+2}+\frac{1}{12}\int\frac{2x-2+6}{x^2-2x+4}dx=\\=-\frac{1}{6}\int\frac{d(x+2)}{x+2}+\frac{1}{12}\int\frac{d(x^2-2x+4)}{x^2-2x+4}+\frac{1}{2}\int\frac{d(x-1)}{(x-1)^2+3}=\\=-\frac{1}{6}ln|x+2|+\frac{1}{12}ln|x^2-2x+4|+\frac{1}{2\sqrt3}arctg\frac{x-1}{\sqrt3}+C$

$\displaystyle\frac{x}{x^3+8}=\frac{A}{x+2}+\frac{Bx+C}{x^2-2x+4}=\frac{-\frac{1}{6}}{x+2}+\frac{\frac{1}{6}x+\frac{1}{3}}{x^2-2x+4}\\x=A(x^2-2x+4)+B(x^2+2x)+C(x+2)\\x^2|0=A+B\\x|1=-2A+2B+C\\x^0|0=4A+2C\\C=\frac{1}{3}\\A=-\frac{1}{6}\\B=\frac{1}{6}$