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