Answer
$$\frac{1}{2}x\sqrt {{x^2} - 25} + \frac{{25}}{2}\ln \left| {\frac{x}{5} + \frac{{\sqrt {{x^2} - 25} }}{5}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2}}}{{\sqrt {{x^2} - 25} }}} dx \cr
& {\text{Substitute }}x = 5\sec \theta ,{\text{ }}dx = 5\sec \theta \tan \theta d\theta \cr
& {\text{Use the trigonometric substitution}} \cr
& \int {\frac{{{x^2}}}{{\sqrt {{x^2} - 25} }}} dx = \int {\frac{{{{\left( {5\sec \theta } \right)}^2}}}{{\sqrt {{{\left( {5\sec \theta } \right)}^2} - 25} }}} \left( {5\sec \theta \tan \theta d\theta } \right) \cr
& {\text{Simplify}} \cr
& = \int {\frac{{25{{\sec }^2}\theta }}{{\sqrt {25{{\sec }^2}\theta - 25} }}} \left( {5\sec \theta \tan \theta d\theta } \right) \cr
& = \int {\frac{{125{{\sec }^2}\theta }}{{5\sqrt {{{\sec }^2}\theta - 1} }}} \left( {\sec \theta \tan \theta d\theta } \right) \cr
& {\text{Where se}}{{\text{c}}^2}\theta - 1 = {\tan ^2}\theta \cr
& = \int {\frac{{25{{\sec }^2}\theta \left( {\sec \theta \tan \theta d\theta } \right)}}{{\sqrt {{{\tan }^2}\theta } }}} \cr
& = \int {\frac{{25{{\sec }^2}\theta \left( {\sec \theta \tan \theta d\theta } \right)}}{{\tan \theta }}} \cr
& = 25\int {{{\sec }^3}\theta } d\theta \cr
& {\text{Use }}{a^{m + n}} = {a^m}{a^n} \cr
& = 25\int {\sec \theta {{\sec }^2}\theta } d\theta \cr
& \cr
& {\text{Integrating by parts}} \cr
& {\text{substitute }}u = \sec \theta ,{\text{ }}du = \sec \theta \tan \theta d\theta \cr
& dv = {\sec ^2}\theta d\theta ,{\text{ }}v = \tan \theta \cr
& \cr
& {\text{Using the formula }}\int {udv} = uv - \int {vdu} {\text{, we have}} \cr
& 25\int {\sec \theta {{\sec }^2}\theta } d\theta = 25\left( {\sec \theta \tan \theta - \int {\tan \theta \left( {\sec \theta \tan \theta d\theta } \right)} } \right) \cr
& = 25\left( {\sec \theta \tan \theta - \int {\sec \theta {{\tan }^2}\theta d\theta } } \right) \cr
& = 25\sec \theta \tan \theta - 25\int {\sec \theta {{\tan }^2}\theta d\theta } \cr
& {\text{where ta}}{{\text{n}}^2}\theta = {\sec ^2}\theta - 1 \cr
& = 25\sec \theta \tan \theta - 25\int {\sec \theta \left( {{{\sec }^2}\theta - 1} \right)d\theta } \cr
& = 25\sec \theta \tan \theta - 25\int {{{\sec }^3}\theta d\theta } + 25\int {\sec \theta } d\theta \cr
& \cr
& {\text{Find the antiderivative by using }} \cr
& \int {{{\sec }^3}\theta } = \frac{1}{2}\sec \theta \tan \theta + \frac{1}{2}\ln \left| {\sec \theta + \tan \theta } \right| + C{\text{ and }}\int {\sec \theta } d\theta = \ln \left| {\sec \theta + \tan \theta } \right| + C \cr
& = 25\sec \theta \tan \theta - 25\left( {\frac{1}{2}\sec \theta \tan \theta + \frac{1}{2}\ln \left| {\sec \theta + \tan \theta } \right|} \right) + 25\ln \left| {\sec \theta + \tan \theta } \right| + C \cr
& = 25\sec \theta \tan \theta - 25\left( {\frac{1}{2}\sec \theta \tan \theta + \frac{1}{2}\ln \left| {\sec \theta + \tan \theta } \right|} \right) + 25\ln \left| {\sec \theta + \tan \theta } \right| + C \cr
& = 25\sec \theta \tan \theta - \frac{{25}}{2}\sec \theta \tan \theta - \frac{{25}}{2}\ln \left| {\sec \theta + \tan \theta } \right| + 25\ln \left| {\sec \theta + \tan \theta } \right| + C \cr
& = \frac{{25}}{2}\sec \theta \tan \theta + \frac{{25}}{2}\ln \left| {\sec \theta + \tan \theta } \right| + C \cr
& \cr
& {\text{Writing the answer in terms of }}x,{\text{ we have that }} \cr
& x = 5\sec \theta ,\,\,\,\,\sqrt {{x^2} - 25} = \sqrt {{{\left( {5\sec \theta } \right)}^2} - 25} = 5\sqrt {{{\tan }^2}\theta } ,\,\,\,\,\tan \theta = \frac{{\sqrt {{x^2} - 25} }}{5} \cr
& {\text{then}}{\text{,}} \cr
& = \frac{{25}}{2}\left( {\frac{x}{5}} \right)\left( {\frac{{\sqrt {{x^2} - 25} }}{5}} \right) + \frac{{25}}{2}\ln \left| {\frac{x}{5} + \frac{{\sqrt {{x^2} - 25} }}{5}} \right| + C \cr
& = \frac{1}{2}x\sqrt {{x^2} - 25} + \frac{{25}}{2}\ln \left| {\frac{x}{5} + \frac{{\sqrt {{x^2} - 25} }}{5}} \right| + C \cr} $$
