Answer
$-\dfrac{1}{x}-\dfrac{x^3}{3}-\dfrac{x}{3}+C$
Work Step by Step
Since, we have $\int (\dfrac{1}{x^2}-x^2-\dfrac{1}{3}) dx=-\dfrac{1}{x}-\dfrac{x^3}{3}-\dfrac{x}{3}+C$
and
$\dfrac{d}{dx}(-\dfrac{1}{x}-\dfrac{x^3}{3}-\dfrac{x}{3}+C)=\dfrac{1}{x^2}-x^2-\dfrac{1}{3}$