Since
$$
\begin{align}
1
&=(\sin^2(x)+\cos^2(x))^2\\[6pt]
&=\sin^4(x)+\cos^4(x)+2\sin^2(x)\cos^2(x)\\[2pt]
&=\sin^4(x)+\cos^4(x)+\frac12\sin^2(2x)
\end{align}
$$
we can proceed with partial fractions
$$
\begin{align}
&\int\frac{\sin(2x)}{\sin^4(x)+\cos^4(x)}\,\mathrm{d}x\\[6pt]
&=\int\frac{\sin(2x)}{1-\frac12\sin^2(2x)}\,\mathrm{d}x\\
&=\frac1{\sqrt2}\int\left(\frac1{1-\frac1{\sqrt2}\sin(2x)}-\frac1{1+\frac1{\sqrt2}\sin(2x)}\right)\,\mathrm{d}x\\
&=\frac1{\sqrt2}\int\left(\frac1{1-\frac{\sqrt2\tan(x)}{1+\tan^2(x)}}-\frac1{1+\frac{\sqrt2\tan(x)}{1+\tan^2(x)}}\right)\frac{\mathrm{d}\tan(x)}{1+\tan^2(x)}\\
&=\frac1{\sqrt2}\int\left(\frac1{\tan^2(x)-\sqrt2\tan(x)+1}-\frac1{\tan^2(x)+\sqrt2\tan(x)+1}\right)\,\mathrm{d}\tan(x)\\
&=\int\left(\frac1{(\sqrt2\tan(x)-1)^2+1}-\frac1{(\sqrt2\tan(x)+1)^2+1}\right)\,\mathrm{d}\sqrt2\tan(x)\\[12pt]
&=\arctan(\sqrt2\tan(x)-1)-\arctan(\sqrt2\tan(x)+1)+C+\tfrac\pi2\\[9pt]
&=\arctan\left(\frac{-1}{\tan^2(x)}\right)+C+\tfrac\pi2\\[9pt]
&=\bbox[5px,border:2px solid #C0A000]{\arctan\left(\tan^2(x)\right)+C}
\end{align}
$$
We can also proceed by writing everything in terms of $\cos(2x)$
$$
\begin{align}
\int\frac{\sin(2x)}{\sin^4(x)+\cos^4(x)}\,\mathrm{d}x
&=\int\frac{\sin(2x)}{1-\frac12\sin^2(2x)}\,\mathrm{d}x\\
&=-\int\frac1{1+\cos^2(2x)}\,\mathrm{d}\cos(2x)\\[9pt]
&=\bbox[5px,border:2px solid #C0A000]{-\arctan(\cos(2x))+C}
\end{align}
$$
These two answers are copacetic since
$$
\begin{align}
\tan\left(\arctan\left(\tan^2(x)\right)+\arctan(\cos(2x))\right)
&=\frac{\tan^2(x)+\cos(2x)}{1-\tan^2(x)\cos(2x)}\\[6pt]
&=\frac{\sin^2(x)+\cos^2(x)(\cos^2(x)-\sin^2(x))}{\cos^2(x)-\sin^2(x)(\cos^2(x)-\sin^2(x))}\\[6pt]
&=\frac{\sin^2(x)+\cos^4(x)-\sin^2(x)\cos^2(x)}{\cos^2(x)+\sin^4(x)-\sin^2(x)\cos^2(x)}\\[6pt]
&=\frac{\sin^4(x)+\cos^4(x)}{\cos^4(x)+\sin^4(x)}\\[12pt]
&=1
\end{align}
$$
That is,
$$
\arctan\left(\tan^2(x)\right)+\arctan\left(\cos(2x)\right)=\frac\pi4
$$