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I have the integer; $\int\frac{\sqrt{16x^2-9}}{x}dx$, 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. $$\int\frac{\sqrt{16x^2-9}}{x}dx$$$$a=3$$$$x=3\sec(\theta)$$ $$dx=3\sec(\theta)\tan(\theta)d\theta$$so $$16x^2-9\rightarrow 16(3\sec(\theta)^2-9$$$$144\sec^2(\theta)-9 \rightarrow 144\left(\sec^2(\theta)-\frac{1}{16}\right)$$ 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.

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Hint:

$$\int\frac{\sqrt{16x^2-9}}{x}dx=\int\frac{\sqrt{(4x)^2-9}}{x}dx$$

you can make $v=4x,dv=4dx$ then you get

$$\int\frac{\sqrt{v^2-9}}{v}dv$$

now you can solve the integral above with the substituition that you maked and back to the variable $x$ making $v=4x$.

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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

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