Answer
$$\frac{x}{{72}}\left( {18{x^2} + 2} \right)\sqrt {9{x^2} + 2} - \frac{1}{{54}}\ln \left| {3x + \sqrt {9{x^2} + 2} } \right| + C$$
Work Step by Step
$$\eqalign{
& \int {{x^2}\sqrt {2 + 9{x^2}} dx} \cr
& {\text{Let }}u = 3x.{\text{ Then }}du = 3dx,{\text{ }}dx = \frac{{du}}{3}{\text{ substituting you have}} \cr
& \int {{x^2}\sqrt {2 + 9{x^2}} dx} = \int {{{\left( {\frac{u}{3}} \right)}^2}\sqrt {2 + {u^2}} \left( {\frac{1}{3}} \right)du} \cr
& = \frac{1}{{27}}\int {{u^2}\sqrt {2 + {u^2}} du} \cr
& {\text{From the table of integrals in the back of the book}} \cr
& \int {{u^2}\sqrt {{u^2} + {a^2}} } du \cr
& = \frac{1}{8}\left[ {u\left( {2{u^2} + {a^2}} \right)\sqrt {{u^2} + {a^2}} - {a^4}\ln \left| {u + \sqrt {{u^2} + {a^2}} } \right|} \right] + C \cr
& \frac{1}{9}\int {{u^2}\sqrt {2 + {u^2}} du} \cr
& = \frac{1}{{216}}\left[ {u\left( {2{u^2} + {{\left( {\sqrt 2 } \right)}^2}} \right)\sqrt {{u^2} + {{\sqrt 2 }^2}} - {{\left( {\sqrt 2 } \right)}^4}\ln \left| {u + \sqrt {{u^2} + {{\sqrt 2 }^2}} } \right|} \right] + C \cr
& = \frac{1}{{216}}\left[ {u\left( {2{u^2} + 2} \right)\sqrt {{u^2} + 2} - 4\ln \left| {u + \sqrt {{u^2} + 2} } \right|} \right] + C \cr
& {\text{Write in terms of }}x,{\text{ }}u = 3x \cr
& = \frac{1}{{216}}\left[ {3x\left( {2{{\left( {3x} \right)}^2} + 2} \right)\sqrt {{{\left( {3x} \right)}^2} + 2} - 4\ln \left| {3x + \sqrt {{{\left( {3x} \right)}^2} + 2} } \right|} \right] + C \cr
& = \frac{1}{{216}}\left[ {3x\left( {18{x^2} + 2} \right)\sqrt {9{x^2} + 2} - 4\ln \left| {3x + \sqrt {9{x^2} + 2} } \right|} \right] + C \cr
& = \frac{{3x}}{{216}}\left( {18{x^2} + 2} \right)\sqrt {9{x^2} + 2} - \frac{4}{{216}}\ln \left| {3x + \sqrt {9{x^2} + 2} } \right| + C \cr
& = \frac{x}{{72}}\left( {18{x^2} + 2} \right)\sqrt {9{x^2} + 2} - \frac{1}{{54}}\ln \left| {3x + \sqrt {9{x^2} + 2} } \right| + C \cr} $$
