Prove the second logarithmic standard form

For $a>0$, write $\frac1{(a-x)(a+x)}=\frac P{a-x}+\frac Q{a+x}$. Multiplying through gives $1=P(a+x)+Q(a-x)$. Setting $x=a$ gives $P=1/(2a)$; setting $x=-a$ gives $Q=1/(2a)$. These substitutions determine the polynomial identity after denominators have been cleared.

$\int\frac1{a^2-x^2}\,dx=\frac1{2a}\int\bigl(\frac1{a-x}+\frac1{a+x}\bigr)\,dx=\frac1{2a}[-\ln|a-x|+\ln|a+x|]+C=\frac1{2a}\ln\left|\frac{a+x}{a-x}\right|+C$, on each interval excluding $x=\pm a$. This supplies the requested proof using the preceding printed partial-fraction argument.