SOLUTION 7: $\ \$ To integrate $\displaystyle{ \int { \sqrt{x^2-9} \over x^2 } \ dx }$ use the trig substitution $$x = 3 \sec \theta$$ so that $$dx = 3 \sec \theta \tan \theta \ d \theta$$ Substitute into the original problem, replacing all forms of x, getting $$\displaystyle { \int \frac{\sqrt{x^{2}-9}}{x^{2}} \ dx = \int \frac{\sqrt{9 \sec^{2} \theta - 9}}{9 \sec^{2} \theta} \cdot 3 \sec \theta \tan \theta \ d \theta }$$ $$= \displaystyle { \int \frac{\sqrt{9 (\sec^{2} \theta - 1 )}}{9 \sec^{2} \theta} \cdot 3 \sec \theta \tan \theta \ d \theta }$$ $$= \displaystyle { \int \frac{\sqrt{9 \tan^{2} \theta}}{9 \sec^{2} \theta} \cdot 3 \sec \theta \tan \theta \ d \theta }$$ $$= \displaystyle { \int \frac{3 \tan \theta}{9 \sec^{2} \theta} \cdot 3 \sec \theta \tan \theta \ d \theta }$$ $$= \displaystyle { \int \frac{\tan^{2} \theta}{\sec \theta} \ d \theta }$$ $$= \displaystyle { \int \frac{\sec^{2} \theta - 1}{\sec \theta} \ d \theta}$$ $$= \displaystyle { \int \Big( \frac{ \sec^2 \theta}{ \sec \theta } - \frac{1}{\sec \theta}) \ d \theta }$$ $$= \displaystyle { \int (\sec \theta - \cos \theta) \ d \theta }$$ (Recall that $\displaystyle { \int \sec \theta \ d \theta = \ln \Big|\sec \theta + \tan \theta \Big| + C } .$) $$= \displaystyle { \ln \Big|\sec \theta + \tan \theta \Big| - \sin \theta + C }$$ $\Big($We need to write our final answer in terms of $x$.

Since $x = 3 \sec \theta$ it follows that $$\sec \theta = \displaystyle{ x \over 3 } = \displaystyle{ hypotenuse \over adjacent }$$ and from the Pythagorean Theorem that $$\displaystyle (adjacent)^2 + (opposite)^2 = (hypotenuse)^2 \ \ \longrightarrow$$ $$(3)^2 + (opposite)^2 = (x)^2 \ \ \longrightarrow \ \ \ opposite = \sqrt{x^2-9} \ \ \longrightarrow$$ $$\sin \theta = \displaystyle{ opposite \over hypotenuse }= \displaystyle{ \sqrt{x^2-9} \over x }$$ and $$\tan \theta = \displaystyle{ opposite \over adjacent }= \displaystyle{ \sqrt{x^2-9} \over 3 } . \Big)$$ $$= \displaystyle { \ln \Big|\frac{x}{3} + \frac{\sqrt{x^{2}-9}}{3} \Big| - \frac{\sqrt{x^{2}-9}}{x} + C }$$

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