<s>[INST] Why do we put absolute brackets for ln? When writing out the final answer in $\ln$ form, why is it necessary to put absolute brackets? How does it affect the answer?
I have this answer of $-3\ln|\frac{3+\sqrt{9-x^2}}{x}|$, but why does it suddenly become 
$3\ln|\frac{\sqrt{9-x^2}-3}{x}|$? 
 [/INST] In this case it's exactly the same:
\begin{align}
\left|\frac{3+\sqrt{9-x^2}}{x}\right|^{\!-1}
&=
\left|\frac{x}{3+\sqrt{9-x^2}}\right|\\[2ex]
&=
\left|\frac{x}{3+\sqrt{9-x^2}}\frac{3-\sqrt{9-x^2}}{3-\sqrt{9-x^2}}\right|\\[2ex]
&=\left|\frac{x(3-\sqrt{9-x^2})}{9-9+x^2}\right|\\[2ex]
&=\left|\frac{3-\sqrt{9-x^2}}{x}\right|\\[2ex]
&=\left|\frac{\sqrt{9-x^2}-3}{x}\right|
\end{align}
Thus
$$
-3\log\left|\frac{3+\sqrt{9-x^2}}{x}\right|=
3\log\left|\frac{3+\sqrt{9-x^2}}{x}\right|^{\!-1}=
3\log\left|\frac{\sqrt{9-x^2}-3}{x}\right|
$$
Why putting the absolute value when writing
$$
\int\frac{1}{x}\,dx=\log|x|+c
$$
and not leaving just $x$? Because this works independently whether the interval where the integral is done is a subset of $(-\infty,0)$ or of $(0,\infty)$. However, one should always recall that such a notation has a meaning only if the integrand is considered defined on an interval.
</s>