<s>[INST] integrate $\int_{2}^{4} \frac{\sqrt{16-x^2}}{x}$ 
$$\int_{2}^{4} \frac{\sqrt{16-x^2}}{x}$$

$x=4\sin(u)$
$dx=4\cos(u)du$
$$\int_{2}^{4} \frac{\sqrt{16-x^2}}{x}=\int_{2}^{4} \frac{\sqrt{16-16\sin^2u}}{4\sin u}\cos u\,du=\int_{2}^{4} \frac{4\sqrt{1-\sin^2u}}{4\sin u}\cos u \,du=\int_{2}^{4} \frac{\cos^2u}{\sin u}du=\int_{2}^{4} \frac{1-\sin^2u}{\sin u}du=\int_{2}^{4} \frac{1}{\sin u}du-\int_{2}^{4} {\sin u}du$$
$$=\ln\left(\tan\left(\frac{u}{2}\right)\right)+\cos u$$
$\frac{x}{4}=\sin u$
$\tan u=\frac{\sqrt{x}}{x^2-16}$
$\ln\left(\frac{{x}}{8(\sqrt{x^2-16})}\right)+\left(\frac{\sqrt{16-x^2}}{4}\right)$ form $2$ to $4$
but I get a $\frac{4}{0}$ the end result according to Wolfram is $1.80$
 [/INST] If we write $$\frac{\sqrt{16-x^2}}{x} = \frac{x \sqrt{16-x^2}}{x^2},$$ then the substitution $u^2 = 16-x^2$, $x^2 = 16-u^2$, $x \, dx = -u \, du$ immediately yields $$\int_{x=2}^4 \frac{\sqrt{16-x^2}}{x} \, dx = \int_{u=\sqrt{12}}^0 \frac{-u^2}{16-u^2} \, du = \int_{u=0}^{\sqrt{12}} \frac{16}{16-u^2} - 1 \, dy.$$  Then we proceed by partial fraction decomposition, giving $$2 \int_{u=0}^{\sqrt{12}} \left(\frac{1}{4-u} + \frac{1}{4+u}\right) \, du - \sqrt{12} = 2 \log \frac{4+\sqrt{12}}{4-\sqrt{12}} - \sqrt{12},$$ which simplifies to $$4 \log (2 + \sqrt{3}) - 2 \sqrt{3}.$$

Since people seem to think that the above calculation is incorrect, here it is via trigonometric substitution along the same lines as attempted by the original question:
$$\begin{align*} \int_{x=2}^4 \frac{\sqrt{16-x^2}}{x} \, dx &= \int_{u = \pi/6}^{\pi/2} \frac{4 \sqrt{1 - \sin^2 u}}{4 \sin u} \cdot 4 \cos u \, du \\ &= 4 \int_{u=\pi/6}^{\pi/2} \frac{\cos^2 u}{\sin u} \, du \\ &= 4 \int_{u=\pi/6}^{\pi/2} \csc u - \sin u \, du \\ &= 4 \int_{u=\pi/6}^{\pi/2} \frac{\csc u (\csc u + \cot u)}{\csc u + \cot u} \, du - \left[ - 4\cos u \right]_{u=\pi/6}^{\pi/2} \\ &= 4\left[-\log|\csc u + \cot u| \right]_{u=\pi/6}^{\pi/2} - 2 \sqrt{3} \\ &=  4\log (2 + \sqrt{3}) - 2 \sqrt{3}.\end{align*}$$

And here is the WolframAlpha link.
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