Answer
$y=\dfrac{-1}{x}+\dfrac{x^2}{2}-\dfrac{1}{2}$
Work Step by Step
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, $y=\dfrac{-1}{x}+\dfrac{x^2}{2}-\dfrac{1}{2}$
