$\int \frac{\sin ^4 x}{\sin ^4 x+\cos ^4 x} d x$ is equal to

$\frac{1}{2}\left\{x-\frac{1}{2 \sqrt{2}} \log \left|\frac{\sqrt{2}+\sin 2 x}{\sqrt{2}-\sin 2 x}\right|\right\}+C$

Let  $I=\int \frac{\sin ^4 x}{\sin ^4 x+\cos ^4 x} d x$. Then,

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

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

$\Rightarrow I=\frac{x}{2}-\frac{1}{2} \int \frac{\cos 2 x}{1-\frac{1}{2} \sin ^2 2 x} d x$

$\Rightarrow I=\frac{x}{2}-\frac{1}{2} \int \frac{1}{(\sqrt{2})^2-(\sin 2 x)^2} d(\sin 2 x)$

$\Rightarrow I=\frac{x}{2}-\frac{1}{4 \sqrt{2}} \log \left|\frac{\sqrt{2}+\sin 2 x}{\sqrt{2}-\sin 2 x}\right|+C$