Answer
$$2\left( {{{\sinh }^{ - 1}}3 - {{\sinh }^{ - 1}}1} \right)$$
Work Step by Step
$$\eqalign{
& \int_{25}^{225} {\frac{{dx}}{{\sqrt {{x^2} + 25x} }}} \cr
& {\text{Applying the hint}} \cr
& \int_{25}^{225} {\frac{{dx}}{{\sqrt {{x^2} + 25x} }}} = \int_{25}^{225} {\frac{{dx}}{{\sqrt x \sqrt {x + 25} }}} \cr
& {\text{Rewrite}} \cr
& = \int_{25}^{225} {\frac{{dx}}{{\sqrt x \sqrt {{{\left( {\sqrt x } \right)}^2} + {{\left( 5 \right)}^2}} }}} \cr
& {\text{Let }}u = \sqrt x \Rightarrow du = \frac{1}{{2\sqrt x }}dx,{\text{ }}\frac{{dx}}{{\sqrt x }} = 2du \cr
& {\text{The new limits of integration are:}} \cr
& x = 225 \Rightarrow u = 15 \cr
& x = 25{\text{ }} \Rightarrow u = 5 \cr
& \int_{25}^{225} {\frac{{dx}}{{\sqrt x \sqrt {{{\left( {\sqrt x } \right)}^2} + {{\left( 5 \right)}^2}} }}} = \int_5^{15} {\frac{{2du}}{{\sqrt {{u^2} + {5^2}} }}} \cr
& {\text{Integrate by using the formula }}\int {\frac{{du}}{{\sqrt {{u^2} + {a^2}} }} = {{\sinh }^{ - 1}}\frac{u}{a} + C} \cr
& \int_5^{15} {\frac{{2du}}{{\sqrt {{u^2} + {5^2}} }}} = 2\left[ {{{\sinh }^{ - 1}}\frac{u}{5}} \right]_5^{15} \cr
& {\text{ }} = 2\left[ {{{\sinh }^{ - 1}}\frac{{15}}{5} - {{\sinh }^{ - 1}}\frac{5}{5}} \right] \cr
& {\text{ }} = 2\left( {{{\sinh }^{ - 1}}3 - {{\sinh }^{ - 1}}1} \right) \cr} $$
