Answer
$$\tan^{-1}(\cos ^{2} x)+C$$
Work Step by Step
\begin{aligned}
\int \frac{\sin 2 x}{1+\cos ^{4} x} d x&=\int \frac{2\sin x\cos x}{1+(\cos ^{2} x)^2} d x
\end{aligned}
Let
$$ u=\cos^2 x \ \ \ \to \ \ du =2\cos x\sin xdx$$
\begin{aligned}
\int \frac{\sin 2 x}{1+\cos ^{4} x} d x&=\int \frac{2\sin x\cos x}{1+(\cos ^{2} x)^2} d x\\
&=\int \frac{du}{1+u^2}\\
&=\tan^{-1}(u)+C\\
&=\tan^{-1}(\cos ^{2} x)+C
\end{aligned}
