Answer
$$10$$
Work Step by Step
$$\eqalign{
& \int_0^1 {{{\left( {2x + 1} \right)}^3}} dx \cr
& u = 2x + 1,{\text{ then }}du = 2dx,{\text{ }}dx = \frac{1}{2}du \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = 1,{\text{ }}u = 3 \cr
& {\text{if }}x = 0,{\text{ }}u = 1 \cr
& {\text{so}} \cr
& \int_0^1 {{{\left( {2x + 1} \right)}^3}} dx = \int_1^3 {{u^3}} \left( {\frac{1}{2}du} \right) \cr
& = \frac{1}{2}\int_1^3 {{u^3}} du \cr
& = \frac{1}{2}\left( {\frac{{{u^4}}}{4}} \right)_1^3 \cr
& = \frac{1}{8}\left( {{u^4}} \right)_1^3 \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = \frac{1}{8}\left( {{{\left( 3 \right)}^4} - {{\left( 1 \right)}^4}} \right) \cr
& = 10 \cr} $$
