# Evaluate the integral? :  int 1/( x^2sqrt(x^2-4) ) dx

Sep 30, 2017

$\int \setminus \frac{1}{{x}^{2} \sqrt{{x}^{2} - 4}} \setminus \mathrm{dx} = \frac{\sqrt{{x}^{2} - 4}}{4 x} + C$

#### Explanation:

We seek:

$I = \int \setminus \frac{1}{{x}^{2} \sqrt{{x}^{2} - 4}} \setminus \mathrm{dx}$

Because of the negative sign let us attempt the following substitution:

$x = 2 \sec \theta \implies {x}^{2} = 4 {\sec}^{2} \theta$

And differentiating wrt $\theta$, we get:

$\frac{\mathrm{dx}}{d \theta} = 2 \sec \theta \tan \theta$

Substituting into the integral we get:

$I = \int \setminus \frac{1}{\left(4 {\sec}^{2} \theta\right) \sqrt{4 {\sec}^{2} \theta - 4}} \setminus 2 \sec \theta \tan \theta \setminus d \theta$

$\setminus \setminus = \frac{1}{2} \setminus \int \setminus \frac{1}{\sec \theta \sqrt{4 \left({\sec}^{2} \theta - 1\right)}} \setminus \tan \theta \setminus d \theta$

$\setminus \setminus = \frac{1}{2} \setminus \int \setminus \frac{1}{\sec \theta \sqrt{4 {\tan}^{2} \theta}} \setminus \tan \theta \setminus d \theta$

$\setminus \setminus = \frac{1}{2} \setminus \int \setminus \frac{1}{\sec \theta \left(2 \tan \theta\right)} \setminus \tan \theta \setminus d \theta$

$\setminus \setminus = \frac{1}{4} \setminus \int \setminus \frac{1}{\sec \theta} \setminus d \theta$

$\setminus \setminus = \frac{1}{4} \setminus \int \setminus \cos \theta \setminus d \theta$

$\setminus \setminus = \frac{1}{4} \setminus \sin \theta + C$ ..... [A]

We now have a simple solution, but we need to be able to restore the substitution; so we need to perform some further trigonometric manipulation. now:

$x = 2 \sec \theta \implies \sec \theta = \frac{x}{2}$
$\therefore \frac{1}{\cos \theta} = \frac{x}{2}$
$\therefore \cos \theta = \frac{2}{x}$

And using the pythagorean identity ${\sin}^{2} A + {\cos}^{2} A \equiv 1$ then:

${\sin}^{2} \theta + {\left(\frac{2}{x}\right)}^{2} = 1$
$\therefore {\sin}^{2} \theta = 1 - \frac{4}{x} ^ 2 = \frac{{x}^{2} - 4}{x} ^ 2$
$\therefore \sin \theta = \sqrt{\frac{{x}^{2} - 4}{x} ^ 2} = \frac{\sqrt{{x}^{2} - 4}}{x}$

And now using this result we can restore the earlier substitution in [A], to give:

$I = \frac{1}{4} \setminus \frac{\sqrt{{x}^{2} - 4}}{x} + C$
$\setminus \setminus = \frac{\sqrt{{x}^{2} - 4}}{4 x} + C$