The value of   $\int \frac{\sin 2x}{{\sin }^{4}x+{\cos }^{4}x}dx$ can be equal to

$$\begin{gathered} I=\int \frac{\sin 2x}{{\sin }^{4}x+{\cos }^{4}x}dx=\int \frac{2\tan x{\sec }^{2}x}{{\tan }^{4}x+1}dx \\ Put {\tan }^{2}x=t\Rightarrow 2\tan x{\sec }^{2}xdx=dt=\int \frac{dt}{t^2+1}={\tan }^{-1}\left(t\right)+C=\frac{\pi }{2}-{\cot }^{-1}\left({\tan }^{2}x\right)+C \end{gathered}$$

$=-{\cot }^{-1}\left({\tan }^{2}x\right)+C_1=-{\cot }^{-1}\left(\frac{1}{{\cot }^{2x}}\right)+C_1={\cot }^{-1}\left({\cot }^{2}x\right)+C_1$

Also  $\cos 2x=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}\Rightarrow \left(\frac{1-\cos 2x}{1+\cos 2x}\right)={\tan }^{2}x$, using these values in given integral

$I=\int \frac{\sin 2x}{{({\cos }^{2}x-{\sin }^{2}x)}^2+2{\sin }^{2}x{\cos }^{2}x}dx=\int \frac{2\sin 2x}{{\left(\cos 2x\right)}^2+1}dx$

$Put \cos 2x=t\Rightarrow -2\sin 2xdx=dt$

$\Rightarrow I=\int \frac{-dt}{t^2+1}=-{\tan }^{-1}t+C_2=-{\tan }^{-1}\left(\cos 2x\right)+C_2$