Answer
$$\frac{1}{3}x^3-x+2\tan^{-1}x+C$$
Work Step by Step
Given $$\int \frac{x^{4}+1}{x^{2}+1} d x$$
By using long division, we get
\begin{align*}
\int \frac{x^{4}+1}{x^{2}+1} d x&=\int( x^{2}-1) d x+\int \frac{2}{x^{2}+1} d x\\
&=\frac{1}{3}x^3-x+2\tan^{-1}x+C
\end{align*}
