$$
\begin{aligned}
& \int \frac{x^4+1}{1+x^6} d x=\int \frac{x^4-x^2+1+x^2}{1+x^6} d x \\
& =\int \frac{x^4-x^2+1}{1+x^6} d x+\int \frac{x^2}{1+x^6} d x \\
& I_1 \quad I_2 \\
& I_1=\int \frac{x^4-x^2+1}{1+\left(x^2\right)^3} d x=\int \frac{x^4-x^2+1}{\left(1+x^2\right)\left(x^4-x^2+1\right)} d x \\
& =\int \frac{1}{1+x^2} d x=\tan ^{-1} x \\
& I_2=\int \frac{x^2}{1+x^6} d x=\int \frac{x^2}{1+\left(x^3\right)^2} d x \\
&
\end{aligned}
$$

Let $x^3=t$
$$
\begin{aligned}
3 x^2 d x & =d t \Rightarrow x^2 d x=\frac{d t}{3} \\
I_2 & =\frac{1}{3} \int \frac{d t}{1+t^2}=\frac{1}{3} \tan ^{-1}(t)
\end{aligned}
$$

Replacing $t$,
$$
\begin{aligned}
& I_2=\frac{1}{3} \tan ^{-1}\left(x^3\right) \\
\therefore \quad & I_1+I_2=\tan ^{-1} x+\frac{1}{3} \tan ^{-1}\left(x^3\right)+C .
\end{aligned}
$$