<s>[INST] 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?
 [/INST] I suggest that you instead split the integral as
$$
\int\frac{\sin x+\cos x}{\sin^4x+\cos^4x}\,dx=\int\frac{\sin x}{(1-\cos^2x)^2+\cos^4x}\,dx+\int\frac{\cos x}{\sin^4x+(1-\sin^2x)^2}\,dx,
$$
and then let $u=\cos x$ and $u=\sin x$ in the respective integral. You will get a (somewhat nasty) integral of a rational function, but it is standard.
Edit
According to your comment (and I agree), you found out that one get integrals like (after expanding the square)
$$
\int\frac{1}{2u^4-2u^2+1}\,du.
$$
Now, you need to factor the denominator, and this is where things get somewhat nasty. I write down the result (in real terms),
$$
2u^4-2u^2+1=2\Bigl(u^2+\sqrt{1+\sqrt{2}}u+\sqrt{2}/2\Bigr)\Bigl(u^2-\sqrt{1+\sqrt{2}}u+\sqrt{2}/2\Bigr).
$$
Thus, there are constants $A$, $B$, $C$ and $D$ such that
$$
\frac{1}{2u^4-2u^2+1}=\frac{Au+B}{u^2+\sqrt{1+\sqrt{2}}u+\sqrt{2}/2}+\frac{Cu+D}{u^2-\sqrt{1+\sqrt{2}}u+\sqrt{2}/2}.
$$
When that is done, you integrate "as usual". You will typically get logarithms and arctans, depending on what the constants $A$, $B$, $C$ and $D$ turn out to be.
The solution by juantheron is (as often) a short cut.
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