[tex]\displaystyle I = \int_{0}^{\frac{\pi}{4}}\dfrac{\sin(4x)}{\cos^4 x+\sin^4 x}\,dx \\ \\ \\ \sin^2 x+\cos^2 x = 1 \Rightarrow \sin^4 x+\cos^4 x+2\sin^2 x\cos^2 x = 1 \Rightarrow\\ \Rightarrow \cos^4x+\sin^4x +\dfrac{(\sin 2x)^2}{2} = 1 \Rightarrow \cos^4x+\sin^4x = 1-\dfrac{(\sin 2x)^2}{2} \\ \\ I = \int_{0}^{\frac{\pi}{4}}\dfrac{2\sin 2x\cos 2x}{1-\dfrac{(\sin 2x)^2}{2}}\,dx\\ \\ \sin 2x = t \Rightarrow 2\cos 2x\,dx= dt \\\\x=0 \Rightarrow t = 0 \\\\ x= \frac{\pi}{4}\Rightarrow t = 1[/tex]
[tex]I = \displaystyle \int_{0}^1\dfrac{t}{1-\dfrac{t^2}{2}}\, dt = \int_{0}^1\dfrac{2t}{2-t^2}\, dt = \\ \\ =-\int_{0}^1\dfrac{(2-t^2)'}{2-t^2}\, dt = -\ln(2-t^2)\Big|_{0}^1 = \\ \\ =-\ln(2-1)+\ln(2+0) = \boxed{\ln 2}[/tex]
