Evaluate $\int \frac{\sqrt{64x^2-256}}{x}\,dx$ QUESTION

Evaluate $$\int \frac{\sqrt{64x^2-256}}{x}\,dx$$

I've tried this problem multiple times and cant seem to find where I made a mistake. If someone could please help explain where I went wrong I would really appreciate it
MY ATTEMPT

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*Typed


$\newcommand{\dd}{\; \mathrm{d}}\int \frac{\sqrt{64x^2-256}}x \dd x \to 
\int \frac{\sqrt{64(x^2-4)}}x \dd x \to
\int \frac{8\sqrt{x^2-4}}x \dd x$
Use $x=a\sec\theta$, $\dd x=a\sec\theta \tan\theta \dd \theta$.
$a=2$ $\to$ $x=2\sec\theta$, $\dd x=2\sec\theta \tan\theta \dd \theta$.
$=\int \frac{8\sqrt{4\sec^2\theta-4}}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta \to
\int \frac{8\sqrt{4(\sec^2\theta-1)}}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta $
$=\int \frac{8\sqrt{4\tan^2\theta}}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta \to
\int \frac{8(2\tan\theta)}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta$
$=\int 16\tan^2\theta \dd \theta \to
16\int\tan^2\theta \dd \theta \to
\underset{\text{trig. formula}}{\underbrace{16(\theta+\tan\theta)+C}}$
$\Rightarrow 16(\tan\theta-\theta)+C = 16\tan\theta-16\theta$
$x=2\sec\theta$, $\sec\theta= \frac x2$
$\boxed{16\tan\left(\frac{\sqrt{x^2-4}}2\right) -16\sec^{-1}\left(\frac x2\right)+C}$


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*Handwritten


 A: Since I am almost blind, I have a lot of problems reading the image.
Consider $$I=\int \frac{\sqrt{64 x^2-256}}{x}\,dx$$ What you apparently did is $x=2\sec(t)$, $dx=2 \tan (t) \sec (t)$ which make $$I=\int \tan (t) \sqrt{256 \sec ^2(t)-256}\,dt=16\int \tan (t) \sqrt{\tan ^2(t)}\,dt=16\int \tan^2 (t) \,dt$$ $$I=16\int (1+\tan^2(t)-1)\,dt=16 (\tan (t)-t)$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With the
  sub$\ds{\ldots\ t \equiv x - \root{x^{2} - 4}\ \imp\
x = {t^{2} + 4 \over 2t}}$:


\begin{align}
&\color{#f00}{\int{\root{64x^{2} - 256} \over x}\,\dd x} =
8\int{\root{x^{2} - 4} \over x}\,\dd x =
8\int\pars{{8 \over t^{2} + 4} - {16 \over t^{2}} + 3}\,\dd t
\\[3mm] = &\
32\arctan\pars{t \over 2} + {128 \over t} + 24t
\\[3mm] = &\ 
32\arctan\pars{x - \root{x^{2} - 4} \over 2} + {128 \over x - \root{x^{2} - 4}} + 24\pars{x - \root{x^{2} - 4}}
\\[3mm] = &\
32\arctan\pars{x - \root{x^{2} - 4} \over 2} +
32\pars{x + \root{x^{2} - 4}} + 24\pars{x - \root{x^{2} - 4}}
\\[3mm] = &\
\color{#f00}{32\arctan\pars{x - \root{x^{2} - 4} \over 2} +
56x + 8\root{x^{2} - 4}} + \pars{~\mbox{a constant}~}
\end{align}
A: $$\dfrac{\sqrt{64x^2-256}}x=8x\cdot\dfrac{\sqrt{x^2-4}}{x^2}$$
Let $\sqrt{x^2-4}=y\implies x^2-4=y^2\implies x\ dx= y\ dy$
$$\int\dfrac{\sqrt{64x^2-256}}xdx=8\int\dfrac{y^2dy}{y^2+4}=8\int dy-32\int\dfrac{dy}{y^2+4}=?$$
A: \begin{align}
&\int \frac{\sqrt{64x^2-256}}{x}\,dx\\
 =& \int \bigg( \frac{8x}{\sqrt{{x^2}-4}}-\frac{32}{x^2\sqrt{1-\frac4{x^2}}}\bigg)\,dx
=\ 8 \sqrt{{x^2}-4}+16\sin^{-1}\frac2x
\end{align}
