Cho hàm số y=f(x) có đạo hàm trên [0;3], thỏa mãn

Đáp án A

Từ giả thiết $\left\{\begin{array}{l}f\left(3-x\right).f\left(x\right)=1\\ f\left(x\right)=\frac{1}{2}\end{array}\right.\Rightarrow f\left(3\right)=2$ .

Ta có:${\left[1+f\left(3-x\right)\right]}^2.f^2\left(x\right)={\left[f\left(x\right)+f\left(3-x\right).f\left(x\right)\right]}^2={\left[f\left(x\right)+1\right]}^2$

+ Tính $I=\underset{0}{\overset{3}{\int }}\frac{x{f}^{\text{'}}\left(x\right)}{{\left[1+f\left(x\right)\right]}^2}dx=-\underset{0}{\overset{3}{\int }}xd\left(\frac{1}{1+f\left(x\right)}\right)=-\frac{x}{1+f\left(x\right)}\left|\begin{array}{l}3\\ 0\end{array}\right.+\underset{0}{\overset{3}{\int }}\frac{1}{1+f\left(x\right)}dx=-1+J$

+ Tính $J=\underset{0}{\overset{3}{\int }}\frac{1}{1+f\left(x\right)}dx\stackrel{t=3-x}{=}-\underset{3}{\overset{0}{\int }}\frac{1}{1+f\left(3-t\right)}dt=\underset{0}{\overset{3}{\int }}\frac{1}{1+f\left(3-t\right)}dt=\underset{0}{\overset{3}{\int }}\frac{1}{1+f\left(3-x\right)}dx$

$\Rightarrow 2J=\underset{0}{\overset{3}{\int }}\frac{1}{1+f\left(x\right)}dx+\underset{0}{\overset{3}{\int }}\frac{1}{1+f\left(3-x\right)}dx=\underset{0}{\overset{3}{\int }}dx\left(f\left(3-x\right).f\left(x\right)=1\right)=3\Rightarrow J=\frac{3}{2}$

Vậy $I=\underset{0}{\overset{3}{\int }}\frac{x{f}^{\text{'}}\left(x\right)}{{\left[1+f\left(3-x\right)\right]}^2.f^2\left(x\right)}dx=\frac{1}{2}$