Answer
$$\frac{{2 \cdot {{10}^{5\sqrt x + 2}}}}{{5\ln 10}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{10}^{5\sqrt x + 2}}}}{{\sqrt x }}} dx \cr
& {\text{set }}u = 5\sqrt x + 2{\text{ then }}\frac{{du}}{{dx}} = 5\left( {\frac{1}{{2\sqrt x }}} \right) \to \frac{1}{{\sqrt x }}dx = \frac{2}{5}du \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{{{{10}^{5\sqrt x + 2}}}}{{\sqrt x }}} dx = \int {{{10}^u}} \left( {\frac{2}{5}du} \right) \cr
& = \frac{2}{5}\int {{{10}^u}} du \cr
& {\text{integrate by the formula }}\int {{a^u}du = \frac{{{a^u}}}{{\ln a}} + C} \cr
& = \frac{2}{5}\left( {\frac{{{{10}^u}}}{{\ln 10}}} \right) + C \cr
& = \frac{{2 \cdot {{10}^u}}}{{5\ln 10}} + C \cr
& {\text{replace }}5\sqrt x + 2{\text{ for }}u \cr
& = \frac{{2 \cdot {{10}^{5\sqrt x + 2}}}}{{5\ln 10}} + C \cr} $$
