Answer
$$\frac{1}{{10}}\left( {t - 2} \right){\left( {4t + 12} \right)^{3/2}} + C$$
Work Step by Step
$$\eqalign{
& \int {t\sqrt {4t + 12} dt} \cr
& {\text{A matching integral in a table of integrals at the end of the book is the }} \cr
& {\text{formula }}93 \cr
& \int {x\sqrt {ax + b} dx = \frac{2}{{15{a^2}}}\left( {3ax - 2b} \right){{\left( {ax + b} \right)}^{3/2}} + C} \cr
& x = t,{\text{ }}a = 4,{\text{ }}b = 12 \cr
& \int {t\sqrt {4t + 12} dt} = \frac{2}{{15{{\left( 4 \right)}^2}}}\left( {3\left( 4 \right)t - 2\left( {12} \right)} \right){\left( {4t + 12} \right)^{3/2}} + C \cr
& \int {t\sqrt {4t + 12} dt} = \frac{2}{{240}}\left( {12t - 24} \right){\left( {4t + 12} \right)^{3/2}} + C \cr
& \int {t\sqrt {4t + 12} dt} = \frac{{24}}{{240}}\left( {t - 2} \right){\left( {4t + 12} \right)^{3/2}} + C \cr
& \int {t\sqrt {4t + 12} dt} = \frac{1}{{10}}\left( {t - 2} \right){\left( {4t + 12} \right)^{3/2}} + C \cr} $$
