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Решите задачу:

$\displaystyle\int\frac{dx}{x^4+x}=\int\frac{dx}{x(x+1)(x^2-x+1)}=\\=\int\frac{dx}{x}-\frac{1}{3}\int\frac{dx}{x+1}-\frac{1}{3}\int\frac{(2x-1)dx}{x^2-x+1}=\\=ln|x|-\frac{1}{3}ln|x+1|-\frac{1}{3}ln|x^2-x+1|+C=ln|x|-\frac{1}{3}ln|x^3+1|+C\\\\\\\frac{1}{x(x+1)(x^2-x+1)}=\frac{A}{x}+\frac{B}{x+1}+\frac{Cx+D}{x^2-x+1}\\1=A(x^3+1)+B(x^3-x^2+x)+(Cx+D)(x^2+x)\\x^3|0=A+B+C\\x^2|0=-B+C+D\\x|0=B+D\\x^0|1=A\\\\D-1=C\\A=1;D=\frac{1}{3};C=-\frac{2}{3};B=-\frac{1}{3}$