## Calculus 8th Edition

$$\displaystyle{\int\frac{\sqrt{x^2-1}}{x^4}dx={\frac{\left(\sqrt{ x^2-1}\right)^3}{3x^3}}+C}\\$$
$\displaystyle{I=\int\frac{\sqrt{x^2-1}}{x^4}dx}\\$ $\displaystyle \left[\begin{array}{ll} x=\sec\theta & x^4=\sec^4\theta \\ & \\ \frac{dx}{d\theta}=\sec\theta\tan\theta & dx=\sec\theta\tan\theta\ d\theta \end{array}\right]$ Integration by substitution $\displaystyle{I=\int\frac{\sqrt{\sec^2\theta-1}}{\sec^4 \theta}\sec\theta\tan\theta\ d\theta}\\ \displaystyle{I=\int\frac{\tan\theta}{\sec^4 \theta}\sec\theta\tan\theta\ d\theta}\\ \displaystyle{I=\int\frac{\tan^2\theta}{\sec^3 \theta}\ d\theta}\\ \displaystyle{I=\int{\sin^2\theta}{\cos \theta}\ d\theta}\\ \displaystyle{I=\int\frac{1}{3}\times3{\sin^2\theta}{\cos \theta}\ d\theta}\\ \displaystyle{I=\frac{1}{3}\int\times3{\sin^2\theta}{\cos \theta}\ d\theta}\\ \displaystyle{I=\frac{1}{3}\sin^3\theta+C}\\$ $\sin\theta=\frac{\sqrt{ x^2-1}}{x}\\$ $\displaystyle{I=\frac{1}{3}{\left(\frac{\sqrt{ x^2-1}}{x}\right)}^3+C}\\ \displaystyle{I={\frac{\left(\sqrt{ x^2-1}\right)^3}{3x^3}}+C}\\$