Chapter 5 - The Integral - 5.5 The Fundamental Theorem of Calculus, Part II - Exercises - Page 263: 34

$3\sqrt {9u^{2}+1}+\sqrt {u^{2}+1}$

$G(x)$ = $\int_{-u}^{3u}\sqrt {x^{2}+1}dx$ = $\int_{0}^{3u}\sqrt {x^{2}+1}dx$ - $\int_{0}^{-u}\sqrt {x^{2}+1}dx$ apply chain rule with FTC then get $G'(x)$ = $3\sqrt {9u^{2}+1}+\sqrt {u^{2}+1}$