Answer
$$\frac{{{x^2}}}{4}\sqrt {2{x^4} + 3} + \frac{3}{{4\sqrt 2 }}\ln \left( {\sqrt 2 {x^2} + \sqrt {2{x^4} + 3} } \right) + C$$
Work Step by Step
$$\eqalign{
& \int {x\sqrt {2{x^4} + 3} } dx,\,\,\,\,u = \sqrt 2 {x^2} \cr
& = \int {x\sqrt {{{\left( {\sqrt 2 {x^2}} \right)}^2} + {{\left( {\sqrt 3 } \right)}^2}} } dx \cr
& {\text{Using the given substitution}} \cr
& u = \sqrt 2 {x^2},\,\,\,\,\,\,\,du = 2\sqrt 2 xdx,\,\,\,\,\,\,dx = \frac{1}{{2\sqrt 2 x}}du \cr
& {\text{write in terms of }}u \cr
& \int {x\sqrt {{{\left( {\sqrt 2 {x^2}} \right)}^2} + {{\left( {\sqrt 3 } \right)}^2}} } dx = \int {x\sqrt {{u^2} + {{\left( {\sqrt 3 } \right)}^2}} } \left( {\frac{1}{{2\sqrt 2 x}}du} \right) \cr
& = \frac{1}{{2\sqrt 2 }}\int {\sqrt {{u^2} + {{\left( {\sqrt 3 } \right)}^2}} } du \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By formula 72}} \cr
& \left( {72} \right):\,\,\,\,\,\,\int {\sqrt {{u^2} + {a^2}} du = \frac{u}{2}\sqrt {{u^2} + {a^2}} + \frac{{{a^2}}}{2}\ln \left( {u + \sqrt {{u^2} + {a^2}} } \right) + C} \cr
& {\text{take }}a = \sqrt 3 \cr
& \frac{1}{{2\sqrt 2 }}\int {\sqrt {{u^2} + {{\left( {\sqrt 3 } \right)}^2}} } du = \frac{u}{{4\sqrt 2 }}\sqrt {{u^2} + {{\left( {\sqrt 3 } \right)}^2}} + \frac{{{{\left( {\sqrt 3 } \right)}^2}}}{{4\sqrt 2 }}\ln \left( {u + \sqrt {{u^2} + {{\left( {\sqrt 3 } \right)}^2}} } \right) + C \cr
& {\text{simplifying}} \cr
& = \frac{u}{{4\sqrt 2 }}\sqrt {{u^2} + 3} + \frac{3}{{4\sqrt 2 }}\ln \left( {u + \sqrt {{u^2} + 3} } \right) + C \cr
& {\text{write in terms of }}x{\text{, and replace }}\sqrt 2 {x^2}{\text{ for }}u \cr
& = \frac{{\sqrt 2 {x^2}}}{{4\sqrt 2 }}\sqrt {{{\left( {\sqrt 2 {x^2}} \right)}^2} + 3} + \frac{3}{{4\sqrt 2 }}\ln \left( {\sqrt 2 {x^2} + \sqrt {{{\left( {\sqrt 2 {x^2}} \right)}^2} + 3} } \right) + C \cr
& = \frac{{{x^2}}}{4}\sqrt {2{x^4} + 3} + \frac{3}{{4\sqrt 2 }}\ln \left( {\sqrt 2 {x^2} + \sqrt {2{x^4} + 3} } \right) + C \cr} $$
