Explanation

(a)
Let $I = \int_{8}^{15}\frac{dx}{(x - 3)\sqrt{x + 1}}$
Put $x = \tan^{2}{\theta\ \Rightarrow \ \theta = \tan^{- 1}\sqrt{x}}$
$dx = 2\tan{\theta\sec^{2}{\theta\ d\theta}}$
$\therefore I = \int_{\tan^{- 1}\sqrt{8}}^{\tan^{- 1}\sqrt{15}}{\frac{2\tan{\theta\sec^{2}\theta}}{(\tan^{2}{\theta - 3)\sqrt{\tan^{2}{\theta + 1}}}}d\theta}$
$= \int_{\tan^{- 1}\sqrt{8}}^{\tan^{- 1}\sqrt{15}}{\frac{2\tan{\theta\sec^{2}\theta}}{(\sec^{2}{\theta - 1 - 3)\sec\theta}}d\theta}$
$= \int_{\tan^{- 1}\sqrt{8}}^{\tan^{- 1}\sqrt{15}}\frac{2\tan{\theta\sec^{2}\theta}}{(\sec^{2}{\theta - 4)\sec\theta}}d\theta$
$= \int_{\tan^{- 1}\sqrt{8}}^{\tan^{- 1}\sqrt{15}}{\frac{2\tan{\theta\sec\theta}}{(\sec^{2}{\theta - 4)}}d\theta}$
$= \left\lbrack \frac{1}{2}\log\frac{(\sec{\theta - 2)}}{(\sec{\theta + 2)}} \right\rbrack_{\tan^{- 1}\sqrt{8}}^{\tan^{- 1}\sqrt{15}}\ \ \ \left\lbrack \because\ \int_{}^{}{\frac{1}{x^{2} - a^{2}}dx = \frac{1}{2a}\log\left( \frac{x - a}{x + a} \right)} \right\rbrack\$
$= \frac{1}{2}\left\lbrack \log\left( \frac{\sec{\tan^{- 1}{\sqrt{15} - 2}}}{\sec{\tan^{- 1}{\sqrt{15} + 2}}} \right) - \log\left( \frac{\sec{\tan^{- 1}{\sqrt{8} - 2}}}{\sec{\tan^{- 1}{\sqrt{8} + 2}}} \right) \right\rbrack$
$= \frac{1}{2}\left\lbrack \log\left( \frac{\sec{\sec^{- 1}{4 - 2}}}{\sec{\sec^{- 1}{4 + 2}}} \right) - \log\left( \frac{\sec{\sec^{- 1}{3 - 2}}}{\sec{\sec^{- 1}{3 + 2}}} \right) \right\rbrack$
$= \frac{1}{2}\left\lbrack \log{\frac{2}{6} - \log\frac{1}{5} = \frac{1}{2}\log\frac{5}{3}} \right\rbrack$