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 $1.80$

I think u forgot to change the limits and also factor of 4 is missing while putting dx term.
Edited answer :
Sorry the previous answer I made a slight mistake in taking limits

So our final answer will look like as seen in the photo above.
As you can see that we get 
$tan(\pi/12)$ in the answer.
By placing value of $tan(\pi/12)$ in different form. Look of the answer can be changed.

So now this is the correct answer. Apologies for slight mistake in limits in the previous answer.
