Answer
$$\frac{{15}}{2} + 8\ln \left( 2 \right)$$
Work Step by Step
$$\eqalign{
& \int_0^3 {\sqrt {{x^2} + 16} } dx \cr
& {\text{From the table of integrals in the back of the book}} \cr
& \int {\sqrt {{u^2} + {a^2}} du} = \frac{1}{2}\left[ {u\sqrt {{u^2} + {a^2}} + {a^2}\ln \left| {u + \sqrt {{u^2} + {a^2}} } \right|} \right] + C \cr
& {\text{Therefore,}} \cr
& \int_0^3 {\sqrt {{x^2} + 16} } dx = \frac{1}{2}\left[ {x\sqrt {{x^2} + 16} + {4^2}\ln \left| {x + \sqrt {{x^2} + 16} } \right|} \right]_0^3 \cr
& = \frac{1}{2}\left[ {x\sqrt {{x^2} + 16} + 16\ln \left| {x + \sqrt {{x^2} + 16} } \right|} \right]_0^3 \cr
& {\text{Evaluating}} \cr
& = \frac{1}{2}\left[ {3\sqrt {{3^2} + 16} + 16\ln \left| {3 + \sqrt {{3^2} + 16} } \right|} \right] - \frac{1}{2}\left[ {0 + 16\ln \left| {0 + \sqrt {0 + 16} } \right|} \right] \cr
& = \frac{1}{2}\left[ {15 + 16\ln \left| 8 \right|} \right] - \frac{1}{2}\left[ {16\ln \left| 4 \right|} \right] \cr
& = \frac{{15}}{2} + 8\ln 8 - 8\ln 4 \cr
& = \frac{{15}}{2} + 8\ln \left( {\frac{8}{4}} \right) \cr
& = \frac{{15}}{2} + 8\ln \left( 2 \right) \cr} $$
