Evaluate: ∫ 1/(sin⁴ x + cos⁴ x) dx

Divide the numerator and denominator by $\cos^4 x$: $I = \int \frac{\sec^4 x}{\tan^4 x + 1}\,dx = \int \frac{(1 + \tan^2 x)\sec^2 x}{\tan^4 x + 1}\,dx$ Put $\tan x = t$, so $\sec^2 x\,dx = dt$: $I = \int \frac{1 + t^2}{t^4 + 1}\,dt = \int \frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}}\,dt$ Put $u = t - \frac1t$, so $du = \left(1 + \frac{1}{t^2}\right)dt$ and $t^2 + \frac{1}{t^2} = u^2 + 2$: $I = \int \frac{du}{u^2 + 2} = \frac{1}{\sqrt2}\tan^{-1}\frac{u}{\sqrt2} + C$ Hence $I = \frac{1}{\sqrt2}\tan^{-1}\left(\frac{\tan^2 x - 1}{\sqrt2\tan x}\right) + C$.