$\int \frac{1}{\sin ^4 x+\cos ^4 x} d x$ equals

$\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\tan 2 x}{\sqrt{2}}\right)+C$

We have,

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

∴  $I=\int \frac{1}{\sin ^4 x+\cos ^4 x} d x$

$\Rightarrow I=2 \int \frac{1}{1+\cos ^2 2 x} d x=2 \int \frac{\sec ^2 2 x}{\sec ^2 2 x+1} d x$

$\Rightarrow I=\int \frac{2 \sec ^2 2 x}{2+\tan ^2 2 x} d x=\int \frac{1}{2+\tan ^2 2 x} d(\tan 2 x)$

$\Rightarrow  I=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\tan 2 x}{\sqrt{2}}\right)+C$