<s>[INST] Compute $I=\int\frac{\sin x+\cos x}{\sin^2x+\cos^4x}\mathrm d x$ 
$$I=\int\frac{\sin x+\cos x}{\sin^2x+\cos^4x}\mathrm d x=?$$

I saw the related problem but that didn't help much. I attempted this question by splitting the numerator into $$I_1=\int\frac{\sin x}{\sin^2x+\cos^4x}\mathrm d x \text { and } I_2=\int\frac{\cos x}{\sin^2x+\cos^4x}\mathrm d x$$
For $I_1$, I set $t=\sin x$ and solved it the straightforward way. For $I_2$, I set $t=\cos x$ and again solved it the straightforward way.
My method got extremely lengthy in the end, while the question is supposed be decently okay. The answer that my book got is:
$$I=\frac 1{2\sqrt3}\ln\left|\frac{\sqrt3+t}{\sqrt3-t}\right|+\arctan t+C$$
for $t=\sin x-\cos x$. I have absolutely no clue how to get to such an easy result.
 [/INST] 
Here is a sketch only that should facilitate leading to the coveted result.  To that end, we proceed.


First, noting that 
$$\sin^2(x)+\cos^4(x)=\cos^2(x)+\sin^4(x)$$
we can write
$$\begin{align}
\int\frac{\sin(x)+\cos(x)}{\sin^2(x)+\cos^4(x)}\,dx&=\int \frac{\cos(x)}{1-\sin^2(x)+\sin^4(x)}\,dx+\int \frac{\sin(x)}{1-\cos^2(x)+\cos^4(x)}\,dx\\\\
&=\left.\left(\int \frac1{v^4-v^2+1}\,dv\right)\right|_{u=\sin(x)}-\left.\left(\int \frac1{u^4-u^2+1}\,du\right)\right|_{u=\cos(x)}
\end{align}$$

Next, we use partial fraction expansion to write 
$$\begin{align}
\frac{1}{x^4-x^2+1}&=\frac{1}{(x^2+\sqrt 3 x +1)(x^2-\sqrt 3 x+1)}\\\\
&=\frac{1}{2\sqrt3}\left(\frac{x+\sqrt3}{x^2+\sqrt 3x+1}-\frac{x-\sqrt3}{x^2-\sqrt 3x+1}\right)\\\\
&=\frac{1}{2\sqrt3}\left(\frac{x+\sqrt3/2+\sqrt3/2}{x^2+\sqrt 3x+1}-\frac{x-\sqrt3/2-\sqrt3/2}{x^2-\sqrt 3x+1}\right)\\\\
&=\frac{1}{4\sqrt3}\left(\frac{2x+\sqrt3}{x^2+\sqrt 3x+1}-\frac{2x-\sqrt3}{x^2-\sqrt 3x+1}\right)\\\\
&+\frac{1}{2}\left(\frac{1}{x^2+\sqrt 3x+1}-\frac{1}{x^2-\sqrt 3x+1}\right)\tag1
\end{align}$$

To finish, observe that the first two terms on the right-hand side of $(1)$ are perfect differentials and integration leads to logarithm terms.  Integration of the last two terms lead to arctangent terms.
This way forward isn't pretty, but it is effective and tractable.
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