Answer
$x-\dfrac{1}{x}-1$
Work Step by Step
Calculate the anti-derivative.
Then $\dfrac{dy}{dx}=\dfrac{x^2+1}{x^2},y(1)=-1$
Now, $y=\int \dfrac{x^2+1}{x^2} dx=\int (1+\dfrac{1}{x^2}) dx= x+\int x^{-2} dx$
We know that $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+C$
where $C$ is a constant of proportionality.
Thus, $y= x+\dfrac{x^{-2+1}}{-2+1}+C=x-\dfrac{1}{x}+C$
Here, $y(1)=-1 \implies C=-1$
Hence, $y=x-\dfrac{1}{x}-1$
