Ta có 
$ \dfrac{x^4 + 1}{x^6 + 1}=  \dfrac{2}{3}.\dfrac{1}{x^2 + 1}+  \dfrac{1}{3}.\dfrac{x^2 + 1}{x^4-x^2 + 1}$ 
$ \dfrac{x^4 + 1}{x^6 + 1}=  \dfrac{2}{3}.\dfrac{1}{x^2 + 1}+  \dfrac{1}{3}.\dfrac{1}{\left ( \dfrac{x}{1-x^2} \right )^2 + 1}.\dfrac{x^2 + 1}{(x^2-1)^2}$ 
Suy ra 
$\int\limits_0^1 {\dfrac{{x^4 + 1}}{{x^6 + 1}}} dx =\int\limits_0^1 \dfrac{2}{3}.\dfrac{1}{x^2 + 1}dx +\int\limits_0^1  \dfrac{1}{3}.\dfrac{1}{\left ( \dfrac{x}{1-x^2} \right )^2 + 1}.d\left ( \dfrac{x}{1-x^2} \right )$ 
$\int\limits_0^1 {\dfrac{{x^4 + 1}}{{x^6 + 1}}} dx =\left[ { \dfrac{2}{3}\arctan x +  \dfrac{1}{3}\arctan\dfrac{x}{1-x^2}} \right]_0^1 $ 
Đến đây bạn tự thay số vào nhé. 

PT 
$\Leftrightarrow \sin 2x−2\sin^2 x-2\sin x \cos x -2\sin x+3\sin x+3\cos x+3=0$ 
$\Leftrightarrow 1−2\sin^2 x+\sin x+3\cos x+2=0$ 
$\Leftrightarrow (\cos x+\sin x)(\cos x-\sin x)+2(\cos x+\sin x)+(\cos x-\sin x)+2=0$ 
$\Leftrightarrow (\cos x+\sin x+2)(\cos x-\sin x+1)=0$ 
$\Leftrightarrow \left[ {\begin{matrix} \cos x+\sin x=-2\\ \cos x+\sin x=-1\end{matrix}} \right.$ 
$\Leftrightarrow \left[ {\begin{matrix} \sin (x+\frac{\pi}{4})=-\sqrt 2   (\text {vô nghiệm})\\ \sin (x+\frac{\pi}{4})=-\frac{1}{\sqrt 2}\end{matrix}} \right.$  
PT còn lại đơn giản bạn tự viết nghiệm nhé.