Answer
$x-\dfrac{1}{x}-1$
Work Step by Step
As we know that $\int x^{a} dx=\dfrac{x^{a+1}}{a+1}+C$
Now, we have $\dfrac{dy}{dx}=\dfrac{x^2+1}{x^2}$ and $y(1)=-1$
This implies that
$y=\int \dfrac{x^2+1}{x^2} dx=\int (1+\dfrac{1}{x^2}) dx$
$ x+[\dfrac{x^{(-2+1)}}{(-2+1)}]+C=x-(\dfrac{1}{x})+C$
Apply the initial condition, we get $y(1)=-1$ Then $C=-1$
Hence, $y=x-(\dfrac{1}{x})-1$