Given, ∫sin 2x dx/(sin⁴ x + cos⁴ x)

= ∫2*sin x * cos x dx/(sin⁴ x + cos⁴ x)

= ∫{(2*sin x * cos x)/cos⁴ x} dx/(sin⁴ x/cos⁴ x + cos⁴ x/cos⁴ x)    {divide by cos⁴ x}

= ∫{(2*sin x)/cos³ x} dx/(tan⁴ x + 1)

= ∫{(2*tan x * sec² x} dx/(tan⁴ x + 1)

Let u = tan x

=> du = sec² x dx

Now, ∫{(2*tan x * sec² x} dx/(tan⁴ x + 1) = ∫2u du/(u⁴ + 1)

Gain let v = u²

=> dv = 2u du

Now, ∫2u du/(u⁴ + 1) = ∫dv/(v⁴ + 1)

                             = tan^-1 v + C

                             = tan^-1 u² + C

                             = tan^-1 (tan² x) + C

So, ∫sin 2x dx/(sin⁴ x + cos⁴ x) = tan^-1 (tan² x) + C