calculus - Integral $int frac{sqrt{16-x^2}}{x} mathrm{d}x$

The problem is $\displaystyle\int\frac{\sqrt{16-x^2}}{x}\mathrm{d}x$. I've attempted to use a trig substitution with $x=4\sin\theta$ and $\mathrm{d}x=4\cos\theta\ \mathrm{d}\theta$. This yields $ \displaystyle 4 \int\frac{\cos^2\theta}{\sin\theta}\mathrm{d}\theta$ and I attempted to substitute $1-\sin^2 \theta$ for the numerator but that did not appear to yield a tractable integral either. (Similar result attempting to substitute a double angle formula.) I attempted to do an integration by parts with $\displaystyle 4\int\frac{\cos\theta}{\sin\theta}\cos\theta\ \mathrm{d}\theta$ and $u=\cos\theta$ and $\displaystyle \mathrm{d}v=\frac{\cos\theta}{\sin\theta}\mathrm{d}\theta$ which gets me $\displaystyle \cos\theta\ln\sin\theta + \int\ln(\sin\theta) \sin\theta\ \mathrm{d} \theta$ and I don't know how to solve that integral either.

Answer

Write the integral as

$$\int{\sqrt{16-x^2}\over x^2}x\,dx$$

then let $u^2=16-x^2$, so that $u\,du=-x\,dx$ and the substitution gives

$$-\int{u\over16-u^2}u\,du=\int\left(1-{16\over16-u^2}\right)\,du$$

Partial fractions should finish things off.