$4\int \frac{\sqrt{a^6+x^8}}{x}dx\text{ is equal to }$

$$\begin{gathered} x^8=t^2-a^8 \\ 8x^{7}dx=2tdt \\ dx=\frac{t}{4x^7}dt \end{gathered}$$

$\text{ Putting }{a}^6+x^8=t^2\text{, we get }I=\int \frac{t^2}{t^2-a^6}dt$

$=t+\frac{a^3}{2}\ln \left|\frac{t-a^3}{t+a^3}\right|+c$

=$\sqrt{a^6+x^8}+\frac{a^3}{2}\ln \left|\frac{\sqrt{a^6+x^8}-a^3}{\sqrt{a^6+x^8}+a^3}\right|+c$