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.
 A: 
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.
A: If you want to do this the way that was suggested using $t=\sin x-\cos x$:
Then $$t^2=s^2-2sc+c^2\implies 4sc=2-2t^2$$
Also, $$t^4=1-4sc+4s^2c^2$$
So putting these results together gives $$4s^2c^2=t^4-2t^2+1$$
Meanwhile, the denominator of the integrand is $$s^2+c^4=s^2+c^2-s^2c^2=1-s^2c^2$$
$$=\frac 34+\frac 12t^2-\frac 14t^4$$
The numerator is just $dt$, so we end up with $$I=\int\frac{4dt}{3+2t^2-t^4}$$
$$=\int\frac{4dt}{(1+t^2)(3-t^2)}$$
This is very easily decomposed into $$\int\frac{1}{t^2+1}+\frac{1}{3-t^2}dt$$
The expected result follows immediately.
A: Since you have provided the hint in the answer $t = \sin x - \cos x$, we're going to use that. Thus, $\mathrm{d}t= (\cos x+\sin x)\mathrm{d}x$
which is already present in the numerator, so no issues there. 
$t^2 = \sin^2x +\cos^2x -\sin2x$
$t^2 = 1 - \sin2x$
thus $\sin2x = 1- t^2$
In the denominator add and subtract $\cos^2x$ $\implies\sin^2x + \cos^2x -\cos^2x + \cos^4x\implies1 - \cos^2x\sin^2x$
The denominator will reduce to $1 - \sin^2x\cos^2x=1 - \sin^22x/4$
Break up the denominator as: $\frac{1}{(1-\sin2x/2)(1+\sin2x/2)}$
now put $\sin2x = 1 - t^2$
Now if you are comfortable with partial fractions the rest should be fairly easy! You can now integrate it with respect to $\mathrm{d}t$. 
