Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.5 Exercises - Page 363: 76

$\frac{ 6^{(x+4)^2}}{2\log 6}+C$

Note that $\log$ denotes the natural logarithm. $\int(x+4)6^{(x+4)^2}dx$ let $(x+4)^2=u$ $du=2(x+4)dx$ $\int(x+4)6^{(x+4)^2}dx$ $=\frac{1}{2} \int 6^udu$ $=\frac{1}{2}(\frac{6^u}{\log 6})+C$ $=\frac{ 6^{(x+4)^2}}{2\log 6}+C$