We know that $$ \int \frac{dx}{a^2-x^2} = \frac{1}{2a}\ln \left| \frac{a+x}{a-x}\right| +C$$ (That absolute value sign is supposed to be longer. I apologize for ignorance on how to make that longer on MathJax computer code.)
While trying to prove this identity, I came to a rather limited proof, shown below.
Assume that $\lvert x\rvert<\lvert a\rvert$. Then, we can redefine $x$ as $x=a \ \mathrm{tanh} \ u$ This is possible since the range of $\mathrm{tanh} \ u$ is from $-1$ to $1$ for all $u$. Thus $dx = a \ \mathrm{sech}^2 \ u \ du$ and $u = \mathrm{artanh} \ \frac{x}{a}$. Using substitution, we have
$$ \int \frac{dx}{a^2-x^2} =\int \frac{a \ \mathrm{sech}^2 \ u}{a^2-a^2 \ \mathrm{tanh}^2 \ u}du =\int \frac{a \ \mathrm{sech}^2 \ u}{a^2(1- \ \mathrm{tanh}^2 \ u)}du = \int \frac{ \mathrm{sech}^2 \ u}{a \ \mathrm{sech}^2 \ u}du = \int \frac{1}{a}du = \frac{u}{a}+C = \frac{1}{a} \mathrm{artanh} \frac{x}{a}+C = \frac{1}{2a}\ln \frac{a+x}{a-x}+C$$
I now have two questions. (1) How does the absolute value appear in the complete integration shown at the beginning? (2) I only partially proved this because I assumed that $\lvert x\rvert<\lvert a\rvert$. How can I prove this integral if $\lvert x\rvert \ge \lvert a\rvert$, other than just differentiating $ \frac{1}{2a}\ln \lvert \frac{a+x}{a-x}\rvert +C$ and seeing that it works?
If you differentiate $\ln x$, then you get $\frac{1}{x}\;(x>0)$, and if you differentiate $\ln (-x)$, then you get $\frac{1}{x}\;(x<0)$. Therefore, if you find the indefinite integral of $\frac{1}{x}$, then $$ \int \frac{1}{x} dx = \frac{1}{|x|}+C. $$ Using partial fraction, $$ \frac{1}{a^2-x^2}=-\frac{1}{(x-a)(x+a)}=-\frac{1}{2a}\left(\frac{1}{x-a}-\frac{1}{x+a}\right) $$, then $$ \int \frac{1}{a^2-x^2} dx = -\frac{1}{2a} (\ln|x-a|-\ln|x+a|)+C. $$