Q.Evaluate $\displaystyle\int \frac{dx}{x^{2}-16}$.

The denominator $x^{2}-16$ is in the form $x^{2}-a^{2}$ with $a=4$, so standard form 1 applies directly. …

Write the denominator as $x^{2}-4^{2}$, matching standard form 1 with $a=4$:

$\int \frac{dx}{x^{2}-16} = \frac{1}{2(4)}\log\left\lvert\frac{x-4}{x+4}\right\rvert+c = \frac{1}{8}\log\left\lvert\frac{x-4}{x+4}\right\rvert+c.$ …

Taking $a=16$ directly instead of recognising $a^{2}=16\Rightarrow a=4$, which would give a wrong coefficient of $\frac{1}{32}$ instead of $\frac{1}{8}$. Also confusing this with for …