<s>[INST] Which integration formula for $\frac{1}{a^2-x^2}$ is correct? In my book integration formula for $\frac{1}{x^2-a^2}$ is given as $\frac{1}{2a}\ln(\frac{x-a}{x+a})$.
From the above formula we can write the formula for integration of $\frac{1}{a^2-x^2}$ as   $-\frac{1}{2a}\ln(\frac{x-a}{x+a}) = \frac{1}{2a}\ln(\frac{x+a}{x-a}) $ [as we know $\ln(\frac{1}{x})=-\ln(x)$]
But, in my book the integration of  $\frac{1}{a^2-x^2}$ is written as $\frac{1}{2a}\ln(\frac{x+a}{{a-x}}) $.
We know that the integration in the second case cannot be $\frac{1}{2a}\ln(\frac{x+a}{x-a}) $ and $\frac{1}{2a}\ln(\frac{x+a}{{a-x}}) $ at the same time. Which formula is the correct one and why?
 [/INST] Neither is correct.
It's not restrictive to assume $a>0$ (for $a=0$ the antiderivative has a different form; for $a<0$ change the intervals below accordingly).
The function $\frac{1}{x^2-a^2}$ is defined for $x\ne\pm a$ and is negative over $(-a,a)$, positive over $(-\infty,-a)$ and $(a,\infty)$.
The sign of $\frac{x-a}{x+a}$ is the same as the sign of $x^2-a^2$ (for $x\ne\pm a$, of course), so it's clear that
$$
\frac{1}{2a}\ln\frac{x-a}{x+a}
$$
cannot be an antiderivative over $(-a,a)$, because it is undefined there.
There's a simple way out. An antiderivative of $1/x$ is $\ln|x|$ (up to an arbitrary constant over $(-\infty,0)$ and an arbitrary constant on $(0,\infty)$, with no connection between each other).
The partial fraction decomposition is
$$
\frac{1}{x^2-a^2}=\frac{1}{2a}\frac{1}{x-a}-\frac{1}{2a}\frac{1}{x+a}
$$
so we can write
$$
\int\frac{1}{x^2-a^2}\,dx=
\frac{1}{2a}(\ln|x-a|-\ln|x+a|)=
\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|
$$
up to arbitrary constants in the intervals $(-\infty,-a)$, $(-a,a)$ and $(a,\infty)$.
Since $1/(a^2-x^2)=-1/(x^2-a^2)$, we have
$$
\int\frac{1}{a^2-x^2}\,dx=
-\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|=
\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|^{-1}=
\frac{1}{2a}\ln\left|\frac{x+a}{x-a}\right|
$$
again up to constants as before.
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