<s>[INST] 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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$\newcommand{\dd}{\; \mathrm{d}}\int \frac{\sqrt{64x^2-256}}x \dd x \to 
\int \frac{\sqrt{64(x^2-4)}}x \dd x \to
\int \frac{8\sqrt{x^2-4}}x \dd x$
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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 [/INST] 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 $$I=\int \tan (t) \sqrt{256 \sec ^2(t)-256}\,dt=16\int \tan (t) \sqrt{\tan ^2(t)}\,dt=16\int \tan^2 (t) \,dt$$ $$I=16\int (1+\tan^2(t)-1)\,dt=16 (\tan (t)-t)$$
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