Answer
$$\frac{\sqrt{x^{2}+4}}{16 x}-\frac{\left(x^{2}+4\right) \sqrt{x^{2}+4}}{48 x^{3}}+C$$
Work Step by Step
Given $$\int \frac{d x}{x^{4} \sqrt{x^{2}+4}}$$
Let $$x=2\tan u\ \ \ \ \ \ \ \ dx=2\sec^2 udu $$
Then
\begin{align*}
\int \frac{d x}{x^{4} \sqrt{x^{2}+4}}&=\int \frac{2\sec^2 udu }{16\tan ^{4} \sqrt{4\tan^{2}u+4}}\\
&=\int \frac{\sec^2 udu }{8\tan ^{4} \sqrt{4\sec^{2}u}}\\
&=\frac{1}{16}\int \frac{\sec udu }{\tan ^{4} }\\
&=\frac{1}{16}\int \cot^3 u \csc udu\\
&=\frac{1}{16}\int (1-\csc^2 u)\cot u \csc udu\\
&=\frac{1}{16} \left( -\csc u+\frac{1}{3}\csc^3u\right)+C\\
&= \frac{\sqrt{x^{2}+4}}{16 x}-\frac{\left(x^{2}+4\right) \sqrt{x^{2}+4}}{48 x^{3}}+C
\end{align*}
