Answer
$$\frac{1}{32} \ln \left|\frac{x-2}{x+2}\right|-\frac{1}{16} \tan ^{-1}\left(\frac{x}{2}\right)+C$$
Work Step by Step
Given
$$
\int \frac{d x}{x^4-16}
$$
Using partial fractions
\begin{aligned}
\frac{1}{\left(x^2+4\right)\left(x+2\right)\left(x-2\right)}&=\frac{a_1x+a_0}{x^2+4}+\frac{a_2}{x+2}+\frac{a_3}{x-2}\\
&= \frac{ \left(a_1x+a_0\right)\left(x+2\right)\left(x-2\right)+a_2\left(x^2+4\right)\left(x-2\right)+a_3\left(x^2+4\right)\left(x+2\right)}{(x^2+4)(x-2)(x+2)}\\
1&=\left(a_1x+a_0\right)\left(x+2\right)\left(x-2\right)+a_2\left(x^2+4\right)\left(x-2\right)+a_3\left(x^2+4\right)\left(x+2\right)
\end{aligned}
At $x=2\to a_3= \frac{1}{32},\ \ x=-2\to a_2= \frac{-1}{32} $, and
$$1=\left(a_1x+a_0\right)\left(x+2\right)\left(x-2\right)+\left(-\frac{1}{32}\right)\left(x^2+4\right)\left(x-2\right)+\frac{1}{32}\left(x^2+4\right)\left(x+2\right)$$
Then
$$1=a_1x^3+a_0x^2+\frac{1}{8}x^2-4a_1x-4a_0+\frac{1}{2}$$
$$1=a_1x^3+x^2\left(a_0+\frac{1}{8}\right)-4a_1x+\left(\frac{1}{2}-4a_0\right) $$
It follows that
$$a_0=-\frac{1}{8},\:a_1=0$$
Hence
\begin{aligned}
\int \frac{d x}{x^4-16} &=\int\left(\frac{1 / 32}{x-2}-\frac{1 / 32}{x+2}-\frac{1 / 8}{x^2+4}\right) d x\\
&=\frac{1}{32} \ln |x-2|-\frac{1}{32} \ln |x+2|-\frac{1}{8} \cdot \frac{1}{2} \tan ^{-1}\left(\frac{x}{2}\right)+C \\
&=\frac{1}{32} \ln \left|\frac{x-2}{x+2}\right|-\frac{1}{16} \tan ^{-1}\left(\frac{x}{2}\right)+C
\end{aligned}
