Answer
$\int\frac{-2x-1}{(x^2+x+1)^3}dx=\frac{1}{2(x^2+x+1)^2}+C$
Work Step by Step
Substitution:
$u=x^2+x+1$
$\frac{du}{dx}=2x+1$
$dx=\frac{1}{2x+1}du$
$\int\frac{-2x-1}{(x^2+x+1)^3}dx=\int\frac{-(2x+1)}{u^3}\frac{1}{2x+1}du=\int-u^{-3}du=-\frac{u^{-2}}{-2}+C=\frac{1}{2(x^2+x+1)^2}+C$
