$$ \int_a^x \frac {dx} {x^4 - c^4} = \frac {1} {4c^3} \ln \left(\frac {x-c} {x+c} \right)_a^x - \frac {1} {2c^3} \tan^{-1} \Bigl(\frac {x} {c} \Bigr)_a^x $$ $$ =\frac {1} {4c^3} \Bigl[ \ln \Bigl(\frac {x-c} {x+c} \Bigr) - \ln \Bigl(\frac {a-c} {a+c} \Bigr) \Bigr]- \frac {1} {2c^3} \Bigl[\tan^{-1} \Bigl(\frac {x} {c} \Bigr)-\tan^{-1} \Bigl(\frac {a} {c} \Bigr)\Bigr]$$

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?

$$ =\frac {1} {4c^3} \left[ \ln \left|\frac {x-c} {x+c} \right| - \ln \left| \frac {a-c} {a+c} \right| \right] $$

  • $\begingroup$ yes sure ,if you know that $a$ could be less then $c$,use absolute value,because logarithm at negative number is undefined,or you can indicate condition $x>c$ or $c>a$ $\endgroup$ – dato datuashvili Aug 8 '12 at 14:43
  • $\begingroup$ @dato, thank you very much for your kind answer!! $\endgroup$ – Tony Aug 8 '12 at 15:23

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.

  • $\begingroup$ Thank you very much for your kind detailed answer!! $\endgroup$ – Tony Aug 8 '12 at 15:22

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