# How to you integrate (x^2)/(x^2 -4) ^(1/2)?

Jun 4, 2015

Let's rewrite it like this:

$\int \frac{{x}^{2}}{\sqrt{{x}^{2} - 4}} \mathrm{dx}$

With this, ${a}^{2} = 4$ so $a = 2$. We have the form:

$\sqrt{{x}^{2} - {a}^{2}}$

which looks like:

$\sqrt{{\left(a \sec \theta\right)}^{2} - {a}^{2}} = \sqrt{{a}^{2} {\sec}^{2} \theta - {a}^{2}} = a \sqrt{{\sec}^{2} \theta - 1} = a \tan \theta = 2 \tan \theta$

So:
Let $x = a \sec \theta = 2 \sec \theta$
$\mathrm{dx} = a \sec \theta \tan \theta d$$\theta = 2 \sec \theta \tan \theta d$$\theta$

$\int {x}^{2} / \left(\sqrt{{x}^{2} - 4}\right) \mathrm{dx} = \int \frac{4 {\sec}^{2} \theta}{\cancel{2 \tan \theta}} \cdot \cancel{2} \sec \theta \cancel{\tan \theta} d$$\theta$

$= \int 4 {\sec}^{3} \theta d$$\theta$

$= 4 \int {\sec}^{2} \theta \sec \theta d$$\theta$

From here, we have a pretty difficult integral... We have to use a trick with integration by parts. Notice how this can be written as:

$= 4 \int \left(1 + {\tan}^{2} \theta\right) \sec \theta d$$\theta$

Let:
$u = \sec \theta$
$v = \tan \theta$
$\mathrm{du} = \sec \theta \tan \theta d$$\theta$
$\mathrm{dv} = {\sec}^{2} \theta d$$\theta$

= 4[overbrace(secthetatantheta)^(u*v) - intoverbrace(tantheta)^(v)overbrace(secthetatanthetadɵ)^(du)]

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

$= 4 \sec \theta \tan \theta - 4 \int {\sec}^{3} \theta - \sec \theta d$$\theta$

$= 4 \sec \theta \tan \theta - 4 \int {\sec}^{3} \theta d$$\theta + 4 \int \sec \theta d$$\theta$

$= 4 \sec \theta \tan \theta + 4 \ln | \tan \theta + \sec \theta | - 4 \int {\sec}^{3} \theta d$$\theta$

where we already are supposed to know that $\int \sec \theta d$$\theta = \ln | \tan \theta + \sec \theta |$. You should have been shown this in class prior to doing this integral.

Now here's the weird part. The whole time we were solving $\int {\sec}^{3} \theta d$$\theta$, so now we can explicitly say that and use that knowledge.

$4 \int {\sec}^{3} \theta d$$\theta = 4 \sec \theta \tan \theta + 4 \ln | \tan \theta + \sec \theta | + C - 4 \int {\sec}^{3} \theta d$$\theta$

$8 \int {\sec}^{3} \theta d$$\theta = 4 \sec \theta \tan \theta + 4 \ln | \tan \theta + \sec \theta | + C$

$\int {\sec}^{3} \theta d$$\theta = \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \ln | \tan \theta + \sec \theta | + C$

where $C$ is now a new constant, $\frac{1}{8} C$, which we don't need to know.

And finally, we can plug things back in.

$\sec \theta = \frac{x}{2}$
$\tan \theta = \frac{\sqrt{{x}^{2} - 4}}{2}$

We get:

$= \frac{x}{4} \frac{\sqrt{{x}^{2} - 4}}{2} + \frac{1}{2} \ln | \frac{\sqrt{{x}^{2} - 4}}{2} + \frac{x}{2} | + C$