Answer
$$
\int \frac{x^{4} d x}{\sqrt{x^{10}-2}} =\frac{1}{5} \ln \left|x^{5}+\sqrt{x^{10}-2}\right|+C,
$$
where $C $ be an arbitrary constant.
Work Step by Step
$$
\int \frac{x^{4} d x}{\sqrt{x^{10}-2}}
$$
If we make the substitution
$$
u=x^{5} , \quad \text { then } \quad d u=5 x^{4} d x,
$$
and we look at the Table of Integrals , we see that the closest entry is number $43 $ with $a^{2}=2$:
\begin{aligned}
\int \frac{x^{4} d x}{\sqrt{x^{10}-2}} &=\int \frac{x^{4} d x}{\sqrt{\left(x^{5}\right)^{2}-2}} \\
&=\frac{1}{5} \int \frac{d u}{\sqrt{u^{2}-2}} \quad\left[\begin{array}{c}
u =x^{5}, \\
d u=5 x^{4} d x
\end{array}\right] \\
& \stackrel{43}{=} \frac{1}{5} \ln \left|u+\sqrt{u^{2}-2}\right|+C \\
&=\frac{1}{5} \ln \left|x^{5}+\sqrt{x^{10}-2}\right|+C
\end{aligned}
