Answer
$$\frac{{{8^{3{x^2} + 1}}}}{{6\ln 8}} + C$$
Work Step by Step
$$\eqalign{
& \int {x{8^{3{x^2} + 1}}} dx \cr
& {\text{set }}u = 3{x^2} + 1{\text{ then }}\frac{{du}}{{dx}} = 6x \to xdx = \frac{1}{6}du \cr
& {\text{write the integrand in terms of }}u \cr
& \int {x{8^{3{x^2} + 1}}} dx = \int {{8^u}} \left( {\frac{1}{6}du} \right) \cr
& = \frac{1}{6}\int {{8^u}} du \cr
& {\text{integrate by the formula }}\int {{a^u}du = \frac{{{a^u}}}{{\ln a}} + C} \cr
& = \frac{1}{6}\left( {\frac{{{8^u}}}{{\ln 8}}} \right) + C \cr
& = \frac{{{8^u}}}{{6\ln 8}} + C \cr
& {\text{replace }}3{x^2} + 1{\text{ for }}u \cr
& = \frac{{{8^{3{x^2} + 1}}}}{{6\ln 8}} + C \cr} $$
