Chapter 7 - Principles Of Integral Evaluation - 7.4 Trigonometric Substitutions - Exercises Set 7.4 - Page 513: 6

$$\frac{{x\sqrt {{x^2} + 5} }}{2} - \frac{5}{2}\ln \left| {x + \sqrt {{x^2} + 5} } \right| + C$$

$$\eqalign{ & \int {\frac{{{x^2}}}{{\sqrt {5 + {x^2}} }}} dx \cr & {\text{substitute }}x = \sqrt 5 \tan \theta ,{\text{ }}dx = \sqrt 5 {\sec ^2}\theta d\theta \cr & = \int {\frac{{{x^2}}}{{\sqrt {5 + {x^2}} }}} dx = \int {\frac{{{{\left( {\sqrt 5 \tan \theta } \right)}^2}}}{{\sqrt {5 + {{\left( {\sqrt 5 \tan \theta } \right)}^2}} }}} \left( {\sqrt 5 {{\sec }^2}\theta d\theta } \right) \cr & = \int {\frac{{5\sqrt 5 {{\tan }^2}\theta {{\sec }^2}\theta }}{{\sqrt {5 + 5{{\tan }^2}\theta } }}} d\theta \cr & = \int {\frac{{5\sqrt 5 {{\tan }^2}\theta {{\sec }^2}\theta }}{{\sqrt {5\left( {1 + {{\tan }^2}\theta } \right)} }}} d\theta \cr & {\text{pythagorean identity}} \cr & = 5\int {\frac{{{{\tan }^2}\theta {{\sec }^2}\theta }}{{\sqrt {{{\sec }^2}\theta } }}} d\theta \cr & = 5\int {\frac{{{{\tan }^2}\theta {{\sec }^2}\theta }}{{\sqrt {{{\sec }^2}\theta } }}} d\theta \cr & = 5\int {{{\tan }^2}\theta \sec \theta } d\theta \cr & {\text{use the identity }}{\tan ^2}\theta + 1 = {\sec ^2}\theta \cr & = 5\int {\left( {{{\sec }^2}\theta - 1} \right)\sec \theta } d\theta \cr & = 5\int {\left( {{{\sec }^3}\theta - \sec \theta } \right)} d\theta \cr & = 5\int {{{\sec }^3}\theta } d\theta - 5\int {\sec \theta } d\theta \cr & {\text{Where }}\int {{{\sec }^3}\theta } d\theta = \frac{1}{2}\ln \left| {\sec \theta + \tan \theta } \right| + \frac{1}{2}\sec \theta \tan \theta + C{\text{ }} \cr & {\text{and }}\int {\sec \theta } d\theta = \ln \left| {\sec \theta + \tan \theta } \right| + C \cr & \cr & = \frac{5}{2}\ln \left| {\sec \theta + \tan \theta } \right| + \frac{5}{2}\sec \theta \tan \theta - 5\ln \left| {\sec \theta + \tan \theta } \right| + C \cr & {\text{simplify}} \cr & = \frac{5}{2}\sec \theta \tan \theta - \frac{5}{2}\ln \left| {\sec \theta + \tan \theta } \right| + C \cr & \cr & {\text{write in terms of }}x \cr & \sec \theta = \frac{{\sqrt {{x^2} + 5} }}{{\sqrt 5 }},\,\,\,\,\tan \theta = \frac{x}{{\sqrt 5 }} \cr & {\text{substituting}} \cr & = \frac{5}{2}\left( {\frac{{\sqrt {{x^2} + 5} }}{{\sqrt 5 }}} \right)\left( {\frac{x}{{\sqrt 5 }}} \right) - \frac{5}{2}\ln \left| {\frac{{\sqrt {{x^2} + 5} }}{{\sqrt 5 }} + \frac{x}{{\sqrt 5 }}} \right| + C \cr & = \frac{{x\sqrt {{x^2} + 5} }}{2} - \frac{5}{2}\ln \left| {x + \sqrt {{x^2} + 5} } \right| + C \cr} $$