How to integrate $\int \frac{1}{\sin^4x + \cos^4 x} \,dx$?

Simplify the denominator in the following way: $$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-\frac{\sin^2(2x)}{2}=\frac{1+\cos^2(2x)}{2}=\frac{2+\tan^2(2x)}{2\sec^2(2x)}$$ Hence, the integral you are dealing with is: $$\int \frac{2\sec^2(2x)}{2+\tan^2(2x)}\,dx$$ I guess the next step is pretty obvious now. ;)

Another approach:

We have $$ \frac{1}{\sin^4x+\cos^4x},\tag1 $$ Multiply $(1)$ by $\dfrac{\tan^4x}{\tan^4x}$ we obtain $$ \frac{\tan^4x}{\sin^4x(1+\tan^4x)}=\frac{\sec^4x}{1+\tan^4x}=\frac{(1+\tan^2x)\sec^2x}{1+\tan^4x}.\tag2 $$ Letting $t=\tan x$, the integral turns out to be $$\eqalign { \int\frac{1+t^2}{1+t^4}\ dt&=\frac12\int\left[\frac{1}{t^2-\sqrt2t+1}+\frac{1}{t^2+\sqrt2t+1}\right]\ dt\\ &=\frac12\int\left[\frac{1}{\left(t-\dfrac1{\sqrt2}\right)^2+\dfrac34}+\frac{1}{\left(t+\dfrac1{\sqrt2}\right)^2+\dfrac34}\right]\ dt.\tag3 } $$ Using substitution $u=\dfrac{\sqrt3}2\left(t-\dfrac1{\sqrt2}\right)$ and $v=\dfrac{\sqrt3}2\left(t+\dfrac1{\sqrt2}\right)$, the integral in $(3)$ can easily be evaluated.

Addendum :

Another way to evaluate $\displaystyle\int\frac{1+t^2}{1+t^4}\ dt$ is dividing the integrand by $\dfrac{t^2}{t^2}$, we obtain $$\eqalign { \int\frac{1+\dfrac1{t^2}}{t^2+\dfrac1{t^2}}\ dt&=\int\frac{1+\dfrac1{t^2}}{\left(t-\dfrac1{t}\right)^2+2}\ dt. } $$ Now let $u=t-\dfrac1{t}\;\Rightarrow\;du=\left(1+\dfrac1{t^2}\right)\ dt$, the integral turns out to be $$ \int\frac{1}{u^2+2}\ du.\tag4 $$ The evaluation of the integral $(4)$ can follow @achillehui's comment.

Convert the exponential powers to multiple angles. From deMoivre's theorem, with $n\in\mathbb{N}$: $$ \begin{align} \left( e^{i \theta} \right)^{n} &= e^{i n\theta} \\ \left( \cos \theta + i \sin \theta \right)^{n} &= \cos n\theta + i \sin n\theta \end{align} $$ These intermediate formulas may help: $$ \begin{align} \cos 2\theta &= \cos^{2} \theta + \sin^{2} \theta \\ \sin 2\theta &= 2 \cos \theta \sin 2 \theta \\ \end{align} $$ Reduce the denominator $$ \sin ^4(x)+\cos ^4(x) = \frac{1}{4} (\cos (4 x)+3) $$ The primitive is $$ \int \frac{1}{\cos^{4}x + \sin^{4}x} \, dx = \int \frac{1}{\cos (4 x)+3} \, dx = \left(4 \sqrt{2}\right)^{-1}\arctan \left(\frac{\tan (2 x)}{\sqrt{2}}\right) $$

Here is a look at the integrand: