Indefinite integral of $\frac{\sqrt{ 25x^2 - 4}}{x}$ Right off the bat i factored a 4 from the radicand to get it into  a form such that i can leverage ${\tan^2(\theta)}$ = ${\sec^2(\theta)} - 1$ 
Then i set $\frac{25}{4}$${x^2}$ = ${\sec^2(\theta)}$
${x}$ = $\frac{2}{5}\sec(\theta)$
${dx}$ = $\frac{2}{5} \sec(\theta)\tan(\theta) \ d \theta$
Now i went ahead and substituted into the integrand, after simplification and evaluation of the new integral we get 
$2\tan\theta - 2\theta$
Now to sub back in terms of x by using a right triangle $2\tan(\theta$) = $\sqrt{ 25x^2 - 4}$ 
(Note: on the right triangle, Opposite = $\sqrt{ 25x^2 - 4}$, Adjacent = $2$, hypotenuse = $5x$)
However for $2\theta$ i made the resubstitution that 
${x}$ = $\frac{2}{5}(\sec\theta)$
so to solve for $\theta$
$\operatorname {arcsec} \frac{5}{2}$${x}$ = ($\theta$)
However that was the wrong resubstitution for ($\theta$), the correct one was
($\theta$) = ${\arccos(\frac{2}{5x})}$
How did $\arccos$ even get into the picture? we already had ${x} = \frac{2}{5}(\sec(\theta))$ so intuitively i just solved for theta. I dont understand how its $\arccos$ and not $\operatorname {arcsec}$, can someone help me make sense of the last part? pls halp
 A: Generally, seeing $\sqrt{x^2 - a^2}$ or something of the like begs for the substitution of $x = a\cos(u)$, since $\cos^2(x) - 1 = \sin^2(x)$. (Or perhaps the hyperbolic version; similar process, I'll leave that calculation up to you if you prefer.) So, notice, by factoring out $\sqrt{25}=5$ from the numerator,
$$\mathcal I := \int \frac{\sqrt{25x^2 - 4}} x dx = 5 \int \frac{\sqrt{x^2 - (2/5)^2}}{x}dx$$
Let $x = 2/5 \cdot \cos(u)$. Then $dx = -2/5 \cdot \sin(u)du$. Notice the radicand becomes
$$x^2 - (2/5)^2 = (2/5)^2 (\cos(u)^2 - 1) = (2 \sin(u) / 5)^2$$
Thus,
$$\mathcal I = 5 \int \frac 2 5 \frac{\sin(u)}{\cos(u)} \cdot \frac{-2}{5} \sin(u)du$$
Some simplification gets us to
$$\mathcal I = \frac{-4}{5} \int \frac{\sin^2(u)}{\cos(u)}du = \frac{-4}{5} \int \frac{1 - \cos^2(u)}{\cos(u)}du = \frac{-4}{5}\left( \int \sec(u)du-\int\cos(u)du \right)$$
These remaining integrals are fairly elementary. You can see this for a list of various integrals. Most relevant,
$$\begin{align}
\int \cos(u)du &= \sin(u) + C\\
\int \sec(u)du &= \ln|\sec(u)+\tan(u)|+C=2 \text{ arctanh} \left( \tan \frac u 2 \right)+C
\end{align}$$
Everything that remains is just back-substitution, simplification, and a bit of algebra, which I'll leave to you.
A: $x = \frac{2}{5} \cosh \theta$ and then you get a linear combination of $\cosh\theta$ and $\mathrm{sech\,}\theta$ to integrate, both of which are standard.
