<s>[INST] Integral of $\int \frac {\sqrt {x^2 - 4}}{x} dx$ I am trying to find $$\int \frac {\sqrt {x^2 - 4}}{x} dx$$
I make $x = 2 \sec\theta$
$$\int \frac {\sqrt {4(\sec^2 \theta - 1)}}{x} dx$$
$$\int \frac {\sqrt {4\tan^2 \theta}}{x} dx$$
$$\int \frac {2\tan \theta}{x} dx$$
From here I am not too sure what to do but I know I shouldn't have x.
$$\int \frac {2\tan \theta}{2 \sec\theta} dx$$
I also know I shouldn't have dx anymore.
$$dx = 2\sec \theta \tan \theta \; \mathrm d\theta$$
$$\int \frac {2\tan \theta}{2 \sec\theta}  2\sec \theta \tan \theta \; \mathrm d\theta$$
$$\int {2\tan^2 \theta}  \; \mathrm d\theta$$
$$2\int {\tan^2 \theta}  \; \mathrm d\theta$$
I have no idea how to find the integral of $\tan^2 \theta$
So I use Wolfram Alpha:
$$\tan \theta - \theta + c$$
Now I need to replace theta with x.
$$x = 2 \sec\theta$$
With same mathmagics I produce
$$ \frac {x}{2} = \sec \theta$$
$$ \theta = \operatorname {arcsec} \left(\frac{x}{2}\right)$$
$$\tan \left(\operatorname {arcsec} \left(\frac{x}{2}\right)\right) - \left(\operatorname {arcsec} \left(\frac{x}{2}\right)\right) + c$$
This is wrong but I am not sure why.
 [/INST] You are correct. First note that you have not carried a factor of $2$, since your integral is $2 \int \tan^2(\theta) d \theta$.
Hence your solution should read $$2 \tan(\text{arsec}(x/2)) - 2 \text{arcsec}(x/2) + c$$
You may want to rewrite your solution to match with the solution in your text. For instance, $$\text{arsec}(x/2)) = \arctan \left( \dfrac{\sqrt{x^2-4}}{2} \right)$$
So why is the above identity true? If we let $\text{arsec}(x/2)) = \theta$, then we get that $\sec(\theta) = x/2$ i.e. $\sec^2(\theta) = \dfrac{x^2}{4}$. We have that $\tan^2(\theta) = \sec^2(\theta) - 1 = \dfrac{x^2}{4} - 1 = \dfrac{x^2-4}{4}$ i.e. $\tan(\theta) = \dfrac{\sqrt{x^2 - 4}}{2}$. Hence, $$\theta = \arctan \left( \dfrac{\sqrt{x^2-4}}{2}\right)$$
Hence, we have the identity, $$\text{arsec}(x/2)) = \arctan \left( \dfrac{\sqrt{x^2-4}}{2} \right)$$
If you use this, then your solution will read $$\sqrt{x^2 - 4} - 2 \, \text{arsec}(x/2)) + c = \sqrt{x^2 - 4} - 2 \, \arctan \left( \dfrac{\sqrt{x^2-4}}{2} \right) + c$$
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