We have,
$$
\int \frac{x^4+1}{x^6+1} d x=A \tan ^{-1} x+B \tan ^{-1} x^3+c
$$
Let $I=\int \frac{x^4+1}{x^6+1} d x=\int \frac{x^4+1+x^2-x^2}{\left(x^2\right)^3+(1)^3} d x$
$$
\begin{aligned}
= & \int \frac{\left(x^4-x^2+1\right)+x^2}{\left(x^4+1\right)\left(x^4+1-x^2\right)} d x \\
& {\left[\because a^3+b^3=(a+b)\left(a^2+b^2-a b\right)\right] } \\
= & \int \frac{d x}{x^2+1}+\int \frac{x^2}{\left(x^2+1\right)\left(x^4+1-x^2\right)} d x \\
= & \int \frac{d x}{x^2+1}+\int \frac{x^2}{x^6+1} d x \\
= & \tan ^{-1} x+\int \frac{x^2}{\left(x^3\right)^2+1} d x
\end{aligned}
$$
Put $x^3=t$
$$
\begin{aligned}
\Rightarrow \quad 3 x^2 d x & =d t \Rightarrow x^2 d x=\frac{1}{3} d t \\
\therefore \quad \quad \quad & =\tan ^{-1} x+\frac{1}{3} \int \frac{d t}{t^2+1} \\
& =\tan ^{-2} x+\frac{1}{3} \tan ^{-1}(t)+c \\
& =\tan ^{-2} x+\frac{1}{3} \tan ^{-1}\left(x^3\right)+c
\end{aligned}
$$
On comparing it with Eq.(i), we get $A=1$ and $B=\frac{1}{3}$.