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
$3\ln|\frac{\sqrt{9-x^2}-3}{x}|$?
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$\begingroup$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.
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