Question

use any method to evaluate the integral.
int\frac{dx}{(x^{2}-9)^{3/2}},x > 3

int\frac{dx}{(x^{2}-9)^{3/2}}=square

Explanation:

Step1: Use trigonometric substitution

Let $x = 3\sec\theta$, then $dx=3\sec\theta\tan\theta d\theta$. Also, $x^{2}-9 = 9\sec^{2}\theta - 9=9(\sec^{2}\theta - 1)=9\tan^{2}\theta$.

Step2: Rewrite the integral

The integral $\int\frac{dx}{(x^{2}-9)^{\frac{3}{2}}}$ becomes $\int\frac{3\sec\theta\tan\theta d\theta}{(9\tan^{2}\theta)^{\frac{3}{2}}}=\int\frac{3\sec\theta\tan\theta d\theta}{27\tan^{3}\theta}=\frac{1}{9}\int\frac{\sec\theta}{\tan^{2}\theta}d\theta$.
Since $\frac{\sec\theta}{\tan^{2}\theta}=\frac{\frac{1}{\cos\theta}}{\frac{\sin^{2}\theta}{\cos^{2}\theta}}=\frac{\cos\theta}{\sin^{2}\theta}$, the integral is $\frac{1}{9}\int\frac{\cos\theta}{\sin^{2}\theta}d\theta$.

Step3: Use substitution again

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

Step4: Integrate

$\frac{1}{9}\int u^{-2}du=\frac{1}{9}\cdot\frac{u^{-2 + 1}}{-2+1}+C=-\frac{1}{9u}+C$.

Step5: Back - substitute

Since $u = \sin\theta$ and $x = 3\sec\theta\Rightarrow\cos\theta=\frac{3}{x}\Rightarrow\sin\theta=\frac{\sqrt{x^{2}-9}}{x}$.
So the integral is $-\frac{x}{9\sqrt{x^{2}-9}}+C$.

Answer:

$-\frac{x}{9\sqrt{x^{2}-9}}+C$