## University Calculus: Early Transcendentals (3rd Edition)

$-\dfrac{1}{x}-\dfrac{x^3}{3}-\dfrac{x}{3}+C$
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 $\dfrac{d}{dx}(-\dfrac{1}{x}-\dfrac{x^3}{3}-\dfrac{x}{3}+C)=\dfrac{1}{x^2}-x^2-\dfrac{1}{3}$