# Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 287: 60

$y=\int_{1}^{x} \sqrt {1+t^2}dt -2$

#### Work Step by Step

Given: $\dfrac{dy}{dx}=\sqrt {1+t^2}$ $\implies \dfrac{d}{dx} [\int_{1}^{x} \sqrt {1+t^2} dt] -2=\sqrt {1+t^2}$ $\implies y(1)=\int_{1}^{1} \sqrt {1+t^2} dt$ $\implies y(1)=-2$ Hence, $y=\int_{1}^{x} \sqrt {1+t^2}dt -2$

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