Cho hàm số $\Large y=f(x)$ có đạo hàm trên [0;3]; $\Large f(3-x).f(x)=

Cho hàm số $\Large y=f(x)$ có đạo hàm trên [0;3]; $\Large f(3-x).f(x)=1,f(x)\ne -1$ với mọi $\Large x\in \left[ 0;3 \right]$ và $\Large f(0)=\dfrac{1}{2}$ . Tính tích phân : $\Large \int\limits_{0}^{3}{\dfrac{x.{f}'(x)}{{{\left[ 1+f(3-x) \right]}^{2}}.{{f}^{2}}(x)}dx}$

• A.

  $\Large 1$

• B.

   $\Large \dfrac{5}{2}$

• C.

  $\Large \dfrac{1}{2}$

• D.

  $\Large \dfrac{3}{2}$

$\Large {{\left( 1+f(3-x) \right)}^{2}}.{{f}^{2}}(x)={{f}^{2}}(x)+2f(3-x).{{f}^{2}}(x)+{{f}^{2}}(3-x){{f}^{2}}(x)$

$\Large ={{f}^{2}}(x)+2f(x)+1={{\left( f(x)+1 \right)}^{2}}$

(Vì $\Large f\left( {3 - x} \right).f\left( x \right) = 1$)

$\Large I=\int\limits_{0}^{3}{\dfrac{x.{f}'(x)}{{{(1+f(x))}^{2}}}dx}$

Đặt $\Large \left\{ \begin{align}  & u=x \\  & dv=\dfrac{{f}'(x)}{{{(1+f(x))}^{2}}}dx \\ \end{align} \right.$ $\Large \Leftrightarrow \left\{ \begin{align}  & du=dx \\  & v=-\dfrac{1}{1+f(x)} \\ \end{align} \right.$ 

$\Large I=\dfrac{-x}{1+f(x)}\left| \begin{align}  & 3 \\  & 0 \\ \end{align} \right.$ $\Large +\int\limits_{0}^{3}{\dfrac{dx}{1+f(x)}=\dfrac{-3}{1+f(3)}+{{I}_{1}}}$

$\Large f(0)=\dfrac{1}{2}\Rightarrow f(3)=2$

Đặt $\Large t=3-x\Rightarrow dt=-dx$

Đổi cận $\Large x=0\Rightarrow t=3$

$\Large x=3\Rightarrow t=0$

$\Large {{I}_{1}}=\int\limits_{0}^{3}{\dfrac{dt}{1+f(3-t)}}$$\Large = \int\limits_0^3 {\dfrac{{dt}}{{1 + \dfrac{1}{{f\left( t \right)}}}}}$ (vì $\Large f(3 - t).f(t) = 1$)$\Large=\int\limits_{0}^{3}{\dfrac{dx}{1+\dfrac{1}{f(x)}}=\int\limits_{0}^{3}{\dfrac{f(x)dx}{1+f(x)}}}$

$\Large 2{{I}_{1}} = \int\limits_0^3 {\dfrac{{dx}}{{1 + f\left( x \right)}}}  + \int\limits_0^3 {\dfrac{{f\left( x \right)dx}}{{1 + f\left( x \right)}}}$$\Large=\int\limits_{0}^{3}{\dfrac{1+f(x)}{1+f(x)}dx=3\Rightarrow {{I}_{1}}=\dfrac{3}{2}}$

Vậy $\Large I=-1+\dfrac{3}{2}=\dfrac{1}{2}$