Chapter 7 - Principles Of Integral Evaluation - Chapter 7 Review Exercises - Page 558: 26

$$\sqrt {1 + 4{x^2}} + \ln \left| {2x} \right| - \ln \left( {\sqrt {4{x^2} + 1} + 1} \right) + C$$

$$\eqalign{ & \int {\frac{{\sqrt {1 + 4{x^2}} }}{x}} dx \cr & {\text{Substitute }}x = \frac{1}{2}\tan \theta ,{\text{ }}dx = \frac{1}{2}{\sec ^2}\theta d\theta \cr & \int {\frac{{\sqrt {1 + 4{x^2}} }}{x}} dx = \int {\frac{{\sqrt {1 + 4{{\left( {\frac{1}{2}\tan \theta } \right)}^2}} }}{{\frac{1}{2}\tan \theta }}} \left( {\frac{1}{2}{{\sec }^2}\theta d\theta } \right) \cr & {\text{Simplify}} \cr & = \int {\frac{{\sqrt {1 + {{\tan }^2}\theta } }}{{\tan \theta }}} \left( {{{\sec }^2}\theta d\theta } \right) \cr & = \int {\frac{{\sqrt {{{\sec }^2}\theta } }}{{\tan \theta }}} \left( {{{\sec }^2}\theta d\theta } \right) \cr & = \int {\frac{{{{\sec }^3}\theta }}{{\tan \theta }}} d\theta \cr & = \int {{{\sec }^3}\theta \cot \theta } d\theta \cr & {\text{By using basic trigonometric identities: sec}}\theta = \frac{1}{{\cos \theta }}{\text{ and }}\cot \theta = \frac{{\cos \theta }}{{\sin \theta }} \cr & = \int {{{\left( {\frac{1}{{\cos \theta }}} \right)}^3}\left( {\frac{{\cos \theta }}{{\sin \theta }}} \right)} d\theta \cr & = \int {\left( {\frac{1}{{{{\cos }^2}\theta }}} \right)\left( {\frac{1}{{\sin \theta }}} \right)} d\theta \cr & = \int {{{\sec }^2}\theta \left( {\frac{1}{{\sin \theta }}} \right)} d\theta \cr & {\text{where se}}{{\text{c}}^2}\theta = {\tan ^2}\theta + 1 \cr & = \int {\left( {{{\tan }^2}\theta + 1} \right)\left( {\frac{1}{{\sin \theta }}} \right)} d\theta \cr & = \int {\left( {\frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} + 1} \right)\left( {\frac{1}{{\sin \theta }}} \right)} d\theta \cr & = \int {\left( {\frac{{\sin \theta }}{{{{\cos }^2}\theta }} + \frac{1}{{\sin \theta }}} \right)} d\theta \cr & = \int {\left( {{{\left( {\cos \theta } \right)}^{ - 2}}\sin \theta + \csc \theta } \right)} d\theta \cr & {\text{integrating}} \cr & = - \frac{{{{\left( {\cos \theta } \right)}^{ - 1}}}}{{ - 1}} - \ln \left| {\csc \theta + \cot \theta } \right| + C \cr & = \frac{1}{{\cos \theta }} - \ln \left| {\csc \theta + \cot \theta } \right| + C \cr & = \sec \theta - \ln \left| {\csc \theta + \cot \theta } \right| + C \cr & \cr & {\text{Writing the answer in terms of }}x,{\text{ we have that }} \cr & x = \frac{1}{2}\tan \theta ,\,\,\,\,\tan \theta = 2x,\,\,\,\sec \theta = \sqrt {1 + {{\tan }^2}\theta } = \sqrt {1 + 4{x^2}} \cr & \,\cot \theta = \frac{1}{{2x}},\,\,\,\,\,\csc \theta = \frac{1}{{\sin \theta }} = \frac{{\sqrt {1 + 4{x^2}} }}{{2x}} \cr & \cr & {\text{then}}{\text{,}} \cr & = \sec \theta - \ln \left| {\csc \theta + \cot \theta } \right| + C \cr & = \sqrt {1 + 4{x^2}} - \ln \left| {\frac{{\sqrt {1 + 4{x^2}} }}{{2x}} + \frac{1}{{2x}}} \right| + C \cr & {\text{Using logarithmic properties}} \cr & = \sqrt {1 + 4{x^2}} - \ln \left| {\frac{{\sqrt {4{x^2} + 1} + 1}}{{2x}}} \right| + C \cr & = \sqrt {1 + 4{x^2}} + \ln \left| {2x} \right| - \ln \left( {\sqrt {4{x^2} + 1} + 1} \right) + C \cr} $$