[Math] How to integrate $\int\frac{\sqrt{16x^2-9}}{x}dx$

Question

I have the integral
$$\int\frac{\sqrt{16x^2-9}}x\,\mathrm{d}x,$$
and I am having trouble doing the trigonometric substitution. So for integrals in the from of $\sqrt{x^2-a^2}$ where $a$ is a constant is by substituting $x$ for $a\sec\theta$. This is where I am having trouble making this substitution. I will post the steps I have taken and someone please point me in the right direction, thanks a lot for the help in advance.
$$\int\frac{\sqrt{16x^2-9}}x\,\mathrm{d}x$$
$$a=3$$
$$x=3\sec\theta$$
$$\mathrm{d}x=3\sec\theta\tan\theta\,\mathrm{d}\theta$$
so
$$16x^2-9\rightarrow 16\left(3\sec\theta^2-9\right)$$
$$144\sec^2\theta-9 \rightarrow 144\left(\sec^2\theta-\frac1{16}\right)$$
and that is where I would normally substitute that into $\tan^2\theta$ but I am not getting how to go about doing that. Thanks again for all the help in advance again.

Answer 1

we have $$\int {\sqrt{16x^2-9}\over x} dx=\int {x\sqrt{16x^2-9}\over x^2} dx$$ now Put $16x^2-9=u^2$ Therefore, we have $$\int {x\sqrt{16x^2-9}\over x^2} dx=\int\frac{u^2}{u^2+9}du$$ I think from here you can do