$\displaystyle{\frac{\sin^42\theta}{\left(\sin^5\theta+\cos^5\theta\right)^2}=16\frac{(\sin\theta\cos\theta)^4}{(1+\sin 2\theta)\left(\sin^4\theta-\sin^3\theta\cos\theta+\sin^2\theta\cos^2\theta-\sin\theta\cos^3\theta+\cos^4\theta\right)^2}}$
$\displaystyle{\frac{\sin^42\theta}{\left(\sin^5\theta+\cos^5\theta\right)^2}=\frac{16}{(1+\sin 2\theta)\left(\tan^2\theta-\tan\theta+1-\cot\theta+\cot^2\theta\right)^2}}$
So $\displaystyle{\displaystyle{\int_0^\frac{\pi}{4}\frac{\ln\cot\theta}{(\sin^5\theta +\cos^5\theta)^2}\sin^42\theta d\theta}=16\int_0^\frac{\pi}{4}\frac{\ln\cot\theta d\theta}{(1+\sin 2\theta)\left(\tan^2\theta-\tan\theta+1-\cot\theta+\cot^2\theta\right)^2}}$
After substitute $u=\cot\theta$ we gonna get $\displaystyle{\int_1^\infty\frac{u^4\ln u}{\left(1+u^5\right)^2}du}$, then continue with by parts method...
$\displaystyle{\int_1^\infty\frac{u^4\ln u}{\left(1+u^5\right)^2}du=\frac{16}{5}\left(-\left[\frac{\ln u}{1+u^5}\right]_1^\infty+\int_1^\infty\frac{du}{u\left(1+u^5\right)}\right)=\frac{16}{5}\int_1^\infty\frac{du}{u\left(1+u^5\right)}}$
But my sense tell me that $\displaystyle{\frac{1}{u\left(1+u^5\right)}=\frac{1}{u}-\frac{1}{5}\left(\frac{1}{1+u}+\frac{4u^3-3u^2+2u-1}{u^4-u^3+u^2-u+1}\right)}$
Finally the answer come out...$\displaystyle{\frac{16}{5}\int_1^\infty\frac{du}{u\left(1+u^5\right)}=\frac{16}{25}\left[\ln\left(\frac{u^5}{1+u^5}\right)\right]_1^\infty=\frac{16}{25}\ln 2}$
โจทย์ "Medium level" ของพี่ passer-by นี่ผมใช้เวลาตั้งครึ่งเดือนนะครับเนี่ย
