1. **State the problem:** Evaluate the integral $$\int \frac{\sqrt{y^2 - 81}}{y} \, dy$$ for $$y > 9$$. 2. **Identify substitution:** Since the integrand contains $$\sqrt{y^2 - 81}$$, a common substitution for expressions of the form $$\sqrt{y^2 - a^2}$$ is $$y = a \sec \theta$$. Here, $$a = 9$$, so we use $$y = 9 \sec \theta$$. 3. **Find $$dy$$:** Differentiate $$y = 9 \sec \theta$$ with respect to $$\theta$$: $$dy = 9 \sec \theta \tan \theta \, d\theta$$ 4. **Rewrite the integral in terms of $$\theta$$:** Calculate $$\sqrt{y^2 - 81}$$: $$\sqrt{(9 \sec \theta)^2 - 81} = \sqrt{81 \sec^2 \theta - 81} = \sqrt{81(\sec^2 \theta - 1)} = 9 \sqrt{\sec^2 \theta - 1}$$ Recall that $$\sec^2 \theta - 1 = \tan^2 \theta$$, so: $$\sqrt{y^2 - 81} = 9 \tan \theta$$ Substitute into the integral: $$\int \frac{\sqrt{y^2 - 81}}{y} \, dy = \int \frac{9 \tan \theta}{9 \sec \theta} \cdot 9 \sec \theta \tan \theta \, d\theta$$ Simplify the fraction inside the integral: $$\frac{9 \tan \theta}{9 \sec \theta} = \frac{\cancel{9} \tan \theta}{\cancel{9} \sec \theta} = \frac{\tan \theta}{\sec \theta}$$ So the integral becomes: $$\int \frac{\tan \theta}{\sec \theta} \cdot 9 \sec \theta \tan \theta \, d\theta = \int 9 \tan \theta \tan \theta \, d\theta = 9 \int \tan^2 \theta \, d\theta$$ 5. **Evaluate $$\int \tan^2 \theta \, d\theta$$:** Recall the identity: $$\tan^2 \theta = \sec^2 \theta - 1$$ So: $$9 \int \tan^2 \theta \, d\theta = 9 \int (\sec^2 \theta - 1) \, d\theta = 9 \left( \int \sec^2 \theta \, d\theta - \int 1 \, d\theta \right)$$ Integrate: $$9 (\tan \theta - \theta) + C$$ 6. **Back-substitute $$\theta$$ in terms of $$y$$:** Since $$y = 9 \sec \theta$$, then: $$\sec \theta = \frac{y}{9}$$ and $$\theta = \sec^{-1} \left( \frac{y}{9} \right)$$ Also, $$\tan \theta = \sqrt{\sec^2 \theta - 1} = \sqrt{\left( \frac{y}{9} \right)^2 - 1} = \frac{\sqrt{y^2 - 81}}{9}$$ Substitute back: $$9 \left( \frac{\sqrt{y^2 - 81}}{9} - \sec^{-1} \left( \frac{y}{9} \right) \right) + C = \sqrt{y^2 - 81} - 9 \sec^{-1} \left( \frac{y}{9} \right) + C$$ **Final answer:** $$\int \frac{\sqrt{y^2 - 81}}{y} \, dy = \sqrt{y^2 - 81} - 9 \sec^{-1} \left( \frac{y}{9} \right) + C$$