Answer
$${3^{\sqrt 2 + 1}}$$
Work Step by Step
$$\eqalign{
& \int_0^3 {\left( {\sqrt 2 + 1} \right){x^{\sqrt 2 }}} dx \cr
& {\text{Take out the constant from the integral:}} \cr
& = \left( {\sqrt 2 + 1} \right)\int_0^3 {{x^{\sqrt 2 }}} dx \cr
& {\text{integrate using the power rule }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C,{\text{ }}where{\text{ }}n \ne - 1.{\text{ }}S{\text{o}}{\text{,}}} \cr
& = \left( {\sqrt 2 + 1} \right)\left( {\frac{{{x^{\sqrt 2 + 1}}}}{{\sqrt 2 + 1}}} \right)_0^3 \cr
& {\text{simplifying}} \cr
& = \left( {{x^{\sqrt 2 + 1}}} \right)_0^3 \cr
& {\text{use fundamental theorem of calculus: }}\cr
& \int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = {3^{\sqrt 2 + 1}} - {0^{\sqrt 2 + 1}} \cr
& = {3^{\sqrt 2 + 1}} \cr} $$
