## Thomas' Calculus 13th Edition

$y=\dfrac{-1}{x}+\dfrac{x^2}{2}-\dfrac{1}{2}$
The anti-derivative for $\dfrac{dy}{dx}=\dfrac{1}{x^2}+x$ is $y=\dfrac{-1}{x}+\dfrac{x^2}{2}+C$ Apply Initial conditions $y(2)=1$.. This implies $\dfrac{-1}{2}+\dfrac{4}{2}+C=1$ or, $C=-\dfrac{1}{2}$ Hence, $y=\dfrac{-1}{x}+\dfrac{x^2}{2}-\dfrac{1}{2}$