Answer
$y=\dfrac{-1}{x}+\dfrac{x^2}{2}-\dfrac{1}{2}$
Work Step by Step
We need to find the anti-derivative for $\dfrac{dy}{dx}=\dfrac{1}{x^2}+x$
Thus, we have: $y=\dfrac{-1}{x}+\dfrac{x^2}{2}+C$
Apply the initial condition $y(2)=1$ in the above equation to solve for $C$.
we get: $\dfrac{-1}{2}+\dfrac{4}{2}+C=1 \implies C=-\dfrac{1}{2}$
Hence, $y=\dfrac{-1}{x}+\dfrac{x^2}{2}-\dfrac{1}{2}$
