Answer
$y=\int_{1}^{x} \sqrt {1+t^2}dt -2$
Work Step by Step
We are given that $\dfrac{dy}{dx}=\sqrt {1+t^2}$
$\dfrac{d}{dx}[\int_{1}^{x} \sqrt {1+t^2} dt -2]=\sqrt {1+x^2}$
This implies that
$y(1)=\int_{1}^{1} \sqrt {1+x^2} dx-2 =-2$
Hence, $y=\int_{1}^{x} \sqrt {1+t^2}dt -2$