Calculus: Early Transcendentals (2nd Edition)

$$\frac{{1 - \sqrt 3 }}{{16}} + \frac{1}{8}\ln \left( {\frac{{2 + \sqrt 3 }}{{\sqrt 3 }}} \right)$$
\eqalign{ & \int_{4/\sqrt 3 }^4 {\frac{{dx}}{{{x^2}\left( {{x^2} - 4} \right)}}} \cr & x = 2\sec \theta ,{\text{ }}dx = 2\sec \theta \tan \theta d\theta \cr & {\text{new limits of integration}} \cr & \theta = {\sec ^{ - 1}}\left( {x/2} \right) \cr & x = 4,{\text{ }}\theta = \pi /3 \cr & x = 4/\sqrt 3 ,{\text{ }}\theta = \pi /6 \cr & {\text{we have}} \cr & \int_{4/\sqrt 3 }^4 {\frac{{dx}}{{{x^2}\left( {{x^2} - 4} \right)}}} = \int_{\pi /6}^{\pi /3} {\frac{{2\sec \theta \tan \theta d\theta }}{{4{{\sec }^2}\theta \left( {4{{\sec }^2}\theta - 4} \right)}}} \cr & = \int_{\pi /6}^{\pi /3} {\frac{{2\sec \theta \tan \theta d\theta }}{{16{{\sec }^2}\theta \left( {{{\sec }^2}\theta - 1} \right)}}} \cr & {\text{pythagorean identity}} \cr & = \int_{\pi /6}^{\pi /3} {\frac{{\sec \theta \tan \theta d\theta }}{{8{{\sec }^2}\theta \left( {{{\tan }^2}\theta } \right)}}} = \frac{1}{8}\int_{\pi /6}^{\pi /3} {\frac{1}{{\sec \theta \tan \theta }}} d\theta \cr & = \frac{1}{8}\int_{\pi /6}^{\pi /3} {\frac{{{{\cos }^2}\theta }}{{\sin \theta }}} d\theta = \frac{1}{8}\int_{\pi /6}^{\pi /3} {\frac{{1 - {{\sin }^2}\theta }}{{\sin \theta }}} d\theta \cr & = \frac{1}{8}\int_{\pi /6}^{\pi /3} {\left( {\csc \theta - \sin \theta } \right)} d\theta \cr & {\text{integrating}} \cr & = \frac{1}{8}\left[ { - \ln \left| {\csc \theta + \cot \theta } \right| + \cos \theta } \right]_{\pi /6}^{\pi /3} \cr & {\text{evaluate limits}} \cr & = \frac{1}{8}\left[ { - \ln \left| {\csc \frac{\pi }{3} + \cot \frac{\pi }{3}} \right| + \cos \frac{\pi }{3}} \right] - \frac{1}{8}\left[ { - \ln \left| {\csc \frac{\pi }{6} + \cot \frac{\pi }{6}} \right| + \cos \frac{\pi }{6}} \right] \cr & {\text{simplify}} \cr & = \frac{1}{8}\left[ { - \ln \left| {\frac{{2\sqrt 3 }}{3} + \frac{{\sqrt 3 }}{3}} \right| + \frac{1}{2}} \right] - \frac{1}{8}\left[ { - \ln \left| {2 + \sqrt 3 } \right| + \frac{{\sqrt 3 }}{2}} \right] \cr & = \frac{1}{8}\left[ { - \ln \left| {\sqrt 3 } \right| + \frac{1}{2}} \right] - \frac{1}{8}\left[ { - \ln \left| {2 + \sqrt 3 } \right| + \frac{{\sqrt 3 }}{2}} \right] \cr & = - \frac{1}{8}\ln \left( {\sqrt 3 } \right) + \frac{1}{{16}} + \frac{1}{8}\ln \left( {2 + \sqrt 3 } \right) - \frac{{\sqrt 3 }}{{16}} \cr & = \frac{{1 - \sqrt 3 }}{{16}} + \frac{1}{8}\ln \left( {\frac{{2 + \sqrt 3 }}{{\sqrt 3 }}} \right) \cr}