13. $int\frac{(x^{2}-16)^{3/2}}{x^{3}}dx$

Explanation:

Step1: Use trig - substitution

Let $x = 4\sec\theta$, then $dx=4\sec\theta\tan\theta d\theta$. First, $(x^{2}-16)^{\frac{3}{2}}=(16\sec^{2}\theta - 16)^{\frac{3}{2}}=(16(\sec^{2}\theta - 1))^{\frac{3}{2}}=(16\tan^{2}\theta)^{\frac{3}{2}} = 64\tan^{3}\theta$. And $x^{3}=64\sec^{3}\theta$. The integral $\int\frac{(x^{2}-16)^{\frac{3}{2}}}{x^{3}}dx=\int\frac{64\tan^{3}\theta}{64\sec^{3}\theta}\cdot4\sec\theta\tan\theta d\theta=\int\frac{4\tan^{4}\theta}{\sec^{2}\theta}d\theta$. Since $\tan^{2}\theta=\sec^{2}\theta - 1$, then $\tan^{4}\theta=(\sec^{2}\theta - 1)^{2}=\sec^{4}\theta-2\sec^{2}\theta + 1$. So the integral becomes $\int\frac{4(\sec^{4}\theta-2\sec^{2}\theta + 1)}{\sec^{2}\theta}d\theta=\int(4\sec^{2}\theta-8 + 4\cos^{2}\theta)d\theta$. Since $\cos^{2}\theta=\frac{1 + \cos(2\theta)}{2}$, the integral is $\int(4\sec^{2}\theta-8 + 4\cdot\frac{1+\cos(2\theta)}{2})d\theta=\int(4\sec^{2}\theta-8 + 2 + 2\cos(2\theta))d\theta=\int(4\sec^{2}\theta-6 + 2\cos(2\theta))d\theta$.

Step2: Integrate term - by - term

$\int(4\sec^{2}\theta-6 + 2\cos(2\theta))d\theta=4\tan\theta-6\theta+\sin(2\theta)+C$. Since $\sin(2\theta) = 2\sin\theta\cos\theta$, and $x = 4\sec\theta\Rightarrow\cos\theta=\frac{4}{x}\Rightarrow\sin\theta=\frac{\sqrt{x^{2}-16}}{x}\Rightarrow\tan\theta=\frac{\sqrt{x^{2}-16}}{4}$. Also, $\theta=\sec^{-1}(\frac{x}{4})$. $4\tan\theta-6\theta+\sin(2\theta)+C = 4\cdot\frac{\sqrt{x^{2}-16}}{4}-6\sec^{-1}(\frac{x}{4})+2\cdot\frac{\sqrt{x^{2}-16}}{x}\cdot\frac{4}{x}+C$. $=\sqrt{x^{2}-16}-6\sec^{-1}(\frac{x}{4})+\frac{8\sqrt{x^{2}-16}}{x^{2}}+C$.

Answer:

$\sqrt{x^{2}-16}-6\sec^{-1}(\frac{x}{4})+\frac{8\sqrt{x^{2}-16}}{x^{2}}+C$