Evaluating $ \int \frac{\sin x+\cos x}{\sin^4x+\cos^4x}\,\text{d}x$. $$
\int \frac{\sin x+\cos x}{\sin^4x+\cos^4x}\,\text{d}x.
$$
What I have tried:
I tried writing denominator as 
$ \sin^4x+\cos^4x =  1-2\sin^2x\cos^2x $ and $ 2\sin^2x\cos^2x = \frac{1}{2}\sin^2(2x) $
so the integral becomes,
$$
\int\frac{\sin x+\cos x}{1-\frac{\sin^2(2x)}{2}}\,\text{d}x.
$$
Anyone, how do I solve this further?

Let $$\displaystyle I = \int\frac{\sin x+\cos x}{\sin^4x +\cos^4 x}dx = \int\frac{\sin x+\cos x}{1-2\sin^2 x\cos^2 x}dx$$
So we get $$I = 2\int\frac{\sin x+\cos x}{2-(\sin 2x)^2}dx = 2\int\frac{\sin x+\cos x}{\left(\sqrt{2}-\sin 2x\right)\cdot \left(\sqrt{2}+\sin 2x\right)}dx$$
So we get $$I = \frac{1}{\sqrt{2}}\int \left[\frac{1}{\sqrt{2}+\sin 2x}+\frac{1}{\sqrt{2}-\sin 2x}\right]\cdot (\sin x+\cos x)dx$$
So we get $$I = \frac{1}{\sqrt{2}}\int \left[\frac{1}{1+\sqrt{2}-(\sin x-\cos x)^2}+\frac{1}{\sqrt{2}-1+(\sin x-\cos x)^2}\right]\cdot(\sin x+\cos x)dx$$ 
Now Put $(\sin x-\cos x) =t\;,$ Then $(\sin x+\cos x)dx = dt$
