Integrating $\int\frac{\sqrt{16x^2-9}}x\,\mathrm{d}x$? I am trying to differentiate from my previous question, but I am having trouble in the finishing steps. I have the integral $\int\dfrac{\sqrt{16x^2-9}}x\,\mathrm{d}x$.
$$v=4x\implies\mathrm{d}v=4\,\mathrm{d}x$$
$$\int\frac{\sqrt{v^2-9}}{v}\,\mathrm{d}v\implies a=3,\quad v=3\sec\theta,\quad\mathrm{d}v=3\sec\theta\tan\theta\,\mathrm{d}\theta.$$
$$v^2-9=9\sec^2\theta-9=9\left(\sec^2\theta-1\right)=9\tan^2\theta.$$
$$\int\frac{3\tan\theta\cdot3\sec\theta\tan\theta}{3\sec\theta}\,\mathrm{d}\theta=3\int\tan^2\theta\,\mathrm{d}\theta.$$
$$3\int\left(\sec^2\theta-1\right)\mathrm{d}\theta=3\left(\tan\theta-\theta\right).$$
And this where I feel I am going wrong. So from my understanding since I used $\sec\theta$ for the substitution the triangle I am using to use this integral has a hypotenuse of $v$ or $4x$ since I initially assigned that value to b and the adjacent side of $\theta$ is $a$ or $3$, and for the missing side  I got $\sqrt{v^2-9}$ or $\sqrt{16x^2-9}$. So for $\tan\theta$ I got $\dfrac{\sqrt{16x^2-9}}3$ and for $\theta$ I got $\sec^{-1}\left(\dfrac{v}3\right)$ or $\cos\left(\dfrac{4x}3\right)$. So my final answer is $$\sqrt{16x^2-9}-3\cos\left(\frac{4x}3\right)$$
but this is wrong so if someone could tell me where I am going wrong it'd be greatly appreciated. Thanks in advance.

I would rather like to try to tackle it with rationalization other than substitution. $$
\begin{aligned}
\int \frac{\sqrt{16 x^{2}-9}}{x} d x &=\int \frac{16 x^{2}-9}{x \sqrt{16 x^{2}-9}} d x \\
&=\frac{1}{16} \int \frac{16 x^{2}-9}{x^{2}} d\left(16 x^{2}-9\right) \\
&=\frac{1}{16} \int\left(16-\frac{9}{x^{2}}\right) d\left(\sqrt{16 x^{2}-9}\right) \\
&=\sqrt{16 x^{2}-9}-3 \arctan \left(\frac{\sqrt{16 x^{2}-9}}{3}\right)+C \\
& (\textrm{ OR }\sqrt{16 x^{2}-9}-3 \arccos \left(\frac{3}{4 x}\right)+C)
\end{aligned}
$$
