Cho hàm số $y=f(x)$ liên tục trên đoạn $\left[0 ...

Ta có:
$I=\int\limits_{1}^{3}{\left[ f\left( 2x-1 \right)+2x \right]}dx=\int\limits_{1}^{3}{2x}dx+\int\limits_{1}^{3}{f\left( 2x-1 \right)}dx=\underset{{}}{\mathop{\underbrace{\int\limits_{1}^{3}{f\left( 2x-1 \right)}dx}_{{{I}_{1}}}}} +8$.
Đặt $t=2x-1\Rightarrow dt=2dx$.
Đổi cận $x=1\Rightarrow t=1; x=3\Rightarrow t=5$.
Suy ra: ${{I}_{1}}=\dfrac{1}{2}\int\limits_{1}^{5}{f\left( t \right)}dt=\dfrac{1}{2}\int\limits_{1}^{5}{f\left( x \right)}dx \left( * \right)$.
Xét $2\cos x.f(1+4\sin x)-\sin 2x.f(3-2\cos 2x)=\sin 4x+4\sin 2x-4\cos x$
Ta có:
Với $1+4\sin x=1\Rightarrow x=0; 1+4\sin x=5\Rightarrow x=\dfrac{\pi }{2}$
Với $3-2\cos 2x=1\Rightarrow x=0$ ; $3-2\cos 2x=5\Rightarrow x=\dfrac{\pi }{2}$.
Suy ra:
$\underbrace{\int\limits_{0}^{\dfrac{\pi }{2}}{2\cos x.f\left( 1+4\sin x \right)} dx}_{{{I}_{2}}}-\underbrace{\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2x.f\left( 3-2\cos 2x \right) }dx}_{{{I}_{3}}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \sin 4x+4\sin 2x-4\cos x \right) dx} \left( 1 \right)$.
Xét ${{I}_{2}}$
Đặt $u=1+4\sin x\Rightarrow du=4\cos xdx\Rightarrow \dfrac{du}{2}=2\cos xdx$.
Đổi cận: $x=0 \Rightarrow u=1 ; x=\dfrac{\pi}{2} \Rightarrow u=5$
Suy ra: ${{I}_{2}}=\int\limits_{1}^{5}{f\left( u \right)\dfrac{du}{2}}=\dfrac{1}{2}\int\limits_{1}^{5}{f\left( x \right)}dx$
Xét ${{I}_{3}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2x.f\left( 3-2\cos 2x \right) }dx$
Đặt $u=3-2\cos 2x\Rightarrow du=4\sin 2xdx\Rightarrow \dfrac{du}{4}=\sin 2xdx$.
Đổi cận: $x=0 \Rightarrow u=1 ; x=\dfrac{\pi}{2} \Rightarrow u=5$
Suy ra: ${{I}_{3}}=\int\limits_{1}^{5}{f\left( u \right)\dfrac{du}{4}}=\dfrac{1}{4}\int\limits_{1}^{5}{f\left( x \right)}dx$.
Thay vào $\left( 1 \right)$ ta được: $\dfrac{1}{2}\int\limits_{1}^{5}{f\left( x \right)}dx+\dfrac{1}{4}\int\limits_{1}^{5}{f\left( x \right)}dx=0\Leftrightarrow \int\limits_{1}^{5}{f\left( x \right)}dx=0$.
Suy ra ${{I}_{1}}=0\Rightarrow I={{I}_{1}}+8=0+8=8$.