Answer
$$\frac{x}{2}\sqrt {{x^2} + 15} + \frac{{15}}{2}\ln \left| {x + \sqrt {{x^2} + 15} } \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\sqrt {{x^2} + 15} } dx \cr
& {\text{or we can write the radical as}} \cr
& = \int {\sqrt {{x^2} + {{\left( {\sqrt {15} } \right)}^2}} } dx \cr
& {\text{integrate by tables using the formulas on the apendix D for this book}} \cr
& {\text{using the formula 15}}:\,\,\,\,\int {\sqrt {{x^2} + {a^2}} dx = \frac{x}{2}\sqrt {{x^2} + {a^2}} + \frac{{{a^2}}}{2} \cdot \ln \left| {x + \sqrt {{x^2} + {a^2}} } \right|} + C \cr
& {\text{in the integral }}\int {\sqrt {{x^2} + {{\left( {\sqrt {15} } \right)}^2}} } dx{\text{ we can see that }}a = \sqrt {15} \cr
& {\text{then}} \cr
& \int {\sqrt {{x^2} + {{\left( {\sqrt {15} } \right)}^2}} } dx = \frac{x}{2}\sqrt {{x^2} + {{\left( {\sqrt {15} } \right)}^2}} + \frac{{{{\left( {\sqrt {15} } \right)}^2}}}{2} \cdot \ln \left| {x + \sqrt {{x^2} + {{\left( {\sqrt {15} } \right)}^2}} } \right| + C \cr
& {\text{simplifying}} \cr
& \int {\sqrt {{x^2} + {{\left( {\sqrt {15} } \right)}^2}} } dx = \frac{x}{2}\sqrt {{x^2} + 15} + \frac{{15}}{2}\ln \left| {x + \sqrt {{x^2} + 15} } \right| + C \cr} $$
