Calculus 8th Edition

$$\displaystyle{\int_2^3\frac{1}{\left(x^2-1\right)^\frac{3}{2}}dx=\frac{2}{\sqrt{3}}-\frac{3}{\sqrt{8}}}$$
$\displaystyle{I=\int_2^3\frac{1}{\left(x^2-1\right)^\frac{3}{2}}dx}\\$ $\displaystyle \left[\begin{array}{ll} x=\sec\theta & x^2=\sec^2\theta \\ & \\ \frac{dx}{d\theta}=\sec\theta\tan\theta & dx=\sec\theta\tan\theta\ d\theta \end{array}\right]$ Integration by substitution $\displaystyle{I=\int_{\frac{\pi}{3}}^{1.23}\frac{1}{\left(\sec^2x-1\right)^\frac{3}{2}}\sec\theta\tan\theta\ d\theta}\\ \displaystyle{I=\int_{\frac{\pi}{3}}^{1.23}\frac{1}{\tan^3\theta}\sec\theta\tan\theta\ d\theta}\\ \displaystyle{I=\int_{\frac{\pi}{3}}^{1.23}\frac{\cos\theta}{\sin^2\theta}\ d\theta}\\ \displaystyle{I=\int_{\frac{\pi}{3}}^{1.23}\cot\theta cosec\theta\ d\theta}\\ \displaystyle{I=\left[-cosec\theta\right]_{\frac{\pi}{3}}^{1.23}}\\ \displaystyle{I=\left[-\frac{1}{\sin\theta}\right]_{\frac{\pi}{3}}^{1.23}}\\$ $\sin\theta=\frac{\sqrt{ x^2-1}}{x}\\$ $\displaystyle{I=\left[-\frac{x}{\sqrt{ x^2-1}}\right]_{2}^{3}+C}\\ \displaystyle{I=-\frac{3}{\sqrt{ 3^2-1}}+\frac{2}{\sqrt{ 2^2-1}}}\\ \displaystyle{I=\frac{2}{\sqrt{3}}-\frac{3}{\sqrt{8}}}$