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

$\dfrac{x}{5}+\dfrac{1}{x^2}+x^2+C$
We have: $\int (\dfrac{1}{5}-\dfrac{2}{x^3}+2x) dx=\dfrac{x}{5}+\dfrac{1}{x^2}+x^2+C$ and $\dfrac{d}{dx}(\dfrac{x}{5}+\dfrac{1}{x^2}+x^2+C)=\dfrac{1}{5}-\dfrac{2}{x^3}+2x$