$\int \frac{x^4+1}{x^6+1} d x$ is equal to

$\tan ^{-1} x+\frac{1}{3} \tan ^{-1} x^3+C$

Let

$I=\int \frac{x^4+1}{x^6+1} d x=\int \frac{\left(x^4-x^2+1\right)+x^2}{x^6+1} d x$

$\Rightarrow I =\int \frac{x^4-x^2+1}{\left(x^2\right)^3+1} d x+\int \frac{x^2}{x^6+1} d x$

$=\int\frac{x^4-x^2+1}{(x^2+1)(x^4-x^2+1)}+\frac{1}{3}\int\frac{d(x^3)}{(x^3)^2+1}$

$=\tan ^{-1} x+\frac{1}{3} \tan ^{-1} x^3+C$