When a is less than c in $ \int_a^b \frac {dx} {x^4 - c^4} $ $$ \int_a^x \frac {dx} {x^4 - c^4} = \frac {1} {4c^3} \ln \left(\frac {x-c} {x+c} \right)_a^x - \frac {1} {2c^3} \tan^{-1} \Bigl(\frac {x} {c} \Bigr)_a^x $$
$$ =\frac {1} {4c^3} \Bigl[ \ln \Bigl(\frac {x-c} {x+c} \Bigr) - \ln \Bigl(\frac {a-c} {a+c} \Bigr) \Bigr]- \frac {1} {2c^3} \Bigl[\tan^{-1} \Bigl(\frac {x} {c} \Bigr)-\tan^{-1} \Bigl(\frac {a} {c} \Bigr)\Bigr]$$
When $x$ is less than $c$, or when $a$ is less than $c$, the number in the natural log becomes negative. Then, should the general answer have absolute sign instead of parenthesis as below?  Is this a more proper or general answer?
$$ =\frac {1} {4c^3} \left[ \ln \left|\frac {x-c} {x+c} \right| - \ln \left| \frac {a-c} {a+c} \right| \right] $$

Yes.  Notice that
$$
\int \frac{dx}{x+c} = \ln \left| x+c \right| + \text{constant}
$$
(where "constant" actually means a piecewise constant function that is constant on $(-\infty,-c)$ and also on $(-c,\infty)$).  Similar remarks apply with $x-c$ in place of $x-c$.  The partial-fraction decomposition of $1/(x^4-c^4)$ includes both $\text{constant}/(x+c)$ and $\text{constant}/(x-c)$, so those two integrals are where these logarithms come from.
