How do you find appropriate trig substitution for $\int \frac{\sqrt{16x^2 - 9}}{x} \, dx$?
In general, if you have $\sqrt{p x^2\pm q}$
- Make the coefficient $x$ equal to $1$ by taking coefficient of $x^2$ out of square root which gives $$\sqrt{px^2\pm q}=\sqrt p\sqrt{x^2\pm \frac{q}{p}}$$
Above expression: $\sqrt{x^2\pm \frac{q}{p}}$ can be changed into the form: $\sqrt{x^2\pm a^2}$ by equating $a=\sqrt{\dfrac{q}{p}}$
Substitute $x=a\sec u$ for the form $\sqrt{x^2-a^2}$ and $x=a\tan u$ for the form $\sqrt{x^2+a^2}$
For this case: $$\sqrt{16x^2-9}=\sqrt{16}\sqrt{x^2-\frac{9}{16}}$$ $$\sqrt{x^2-a^2}=\sqrt{x^2-\frac{9}{16}}$$ $$\implies a=\sqrt{\frac{9}{16}}=\frac34$$
Note: $\sqrt{16x^2-3^2}$ is a difference of squares. Draw a picture of a right triangle suggested by this: $4x$ the hypotenuse, $3$ one of the legs (say the side opposite angle $\theta$), and $\sqrt{16x^2-3^2}$ the side adjacent to angle $\theta$.
Do it, don't just rely on my description.
So then: $$ \sin\theta = \frac{3}{4x}, \\ \cos\theta = \frac{\sqrt{16x^2-3^2}}{4x}, \\ \tan\theta = \frac{3}{\sqrt{16x^2-3^2}}. $$ Use the simplest one to suggest the substitution: $$ x = \frac{3}{4}\csc \theta, \\ dx = -\frac{3}{4}\csc\theta\cot\theta\;d\theta $$ Then substitute back into your integral, looking at your picture to find how to move between $x$ and $\theta$. Here $$ \frac{\sqrt{16x^2 - 9}}{x} = 4\cos \theta $$ so we get \begin{align} \int\frac{\sqrt{16x^2 - 9}}{x}\;dx &= -\int 4\cos \theta \frac{3}{4}\csc\theta\cot\theta\;d\theta \\ &= -3\int\frac{\cos^2\theta}{\sin^2\theta}\;d\theta = 3\big(\cot \theta + \theta\big)+C \end{align} and then look at the picture to get $$ 3\big(\cot \theta + \theta\big)+C= 3 \left[\frac{\sqrt{16x^2-3^2}}{3} + \arcsin\frac{3}{4x}\right]+C $$
this method also works for "sum of squares". Draw the right triangle suggested by that particular sum of squares.
$$ 16x^2 - 9 = 9\left( \left( \tfrac{4x}{3} \right)^2 - 1 \right) = 9(\sec^2\theta - 1) = 9\tan^2\theta. $$