evaluate $int\frac{sqrt{x^{2}-9}}{x^{4}}dx$.

Answer:

$-\frac{(x^{2}-9)^{\frac{3}{2}}}{27x^{3}}+C$

Explanation:

Step1: Use trig - substitution

Let $x = 3\sec\theta$, then $dx=3\sec\theta\tan\theta d\theta$.

Step2: Substitute into the integral

$\int\frac{\sqrt{x^{2}-9}}{x^{4}}dx=\int\frac{\sqrt{9\sec^{2}\theta - 9}}{81\sec^{4}\theta}\cdot3\sec\theta\tan\theta d\theta$. Since $\sqrt{9\sec^{2}\theta - 9}=3\tan\theta$, the integral becomes $\int\frac{3\tan\theta}{81\sec^{4}\theta}\cdot3\sec\theta\tan\theta d\theta=\frac{1}{9}\int\frac{\tan^{2}\theta}{\sec^{3}\theta}d\theta$.

Step3: Rewrite in terms of sine and cosine

$\frac{1}{9}\int\frac{\sin^{2}\theta}{\cos^{2}\theta}\cdot\cos^{3}\theta d\theta=\frac{1}{9}\int\sin^{2}\theta\cos\theta d\theta$.

Step4: Use substitution

Let $u = \sin\theta$, then $du=\cos\theta d\theta$. The integral is $\frac{1}{9}\int u^{2}du$.

Step5: Integrate

$\frac{1}{9}\cdot\frac{u^{3}}{3}+C=\frac{1}{27}\sin^{3}\theta + C$.

Step6: Back - substitute

Since $x = 3\sec\theta$, $\cos\theta=\frac{3}{x}$ and $\sin\theta=\frac{\sqrt{x^{2}-9}}{x}$. So the result is $-\frac{(x^{2}-9)^{\frac{3}{2}}}{27x^{3}}+C$.