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

Hướng dẫn giải:

Đáp án C.

Ta có:  \( f(3-x).f(x)=1\Rightarrow f(3-x)=\frac{1}{f(x)} \) .

+ Xét  \( {{\left[ 1+f(3-x) \right]}^{2}}.{{f}^{2}}(x)={{f}^{2}}(x)+2.f(3-x).{{f}^{2}}(x)+{{f}^{2}}(3-x).{{f}^{2}}(x) \)

 \( ={{f}^{2}}(x)+2.f(x)+1={{\left[ f(x)+1 \right]}^{2}} \).

+ Xét  \( I=\int\limits_{0}^{3}{\frac{x.{f}'(x)}{{{\left[ 1+f(3-x) \right]}^{2}}.{{f}^{2}}(x)}dx}=\int\limits_{0}^{3}{\frac{x.{f}'(x)}{{{\left[ 1+f(x) \right]}^{2}}}dx} \) .

Đặt  \( \left\{ \begin{align}  & u=x\Rightarrow du=dx \\  & dv=\frac{{f}'(x)}{{{\left[ 1+f(x) \right]}^{2}}}dx=\frac{1}{{{\left[ 1+f(x)\right]}^{2}}}d\left( f(x) \right)\Rightarrow v=-\frac{1}{1+f(x)} \\ \end{align} \right. \).

Khi đó:  \( I=\left. \frac{-x}{1+f(x)} \right|_{0}^{3}+\underbrace{\int\limits_{0}^{3}{\frac{1}{1+f(x)}dx}}_{{{I}_{1}}}=\frac{-3}{1+f(3)}+{{I}_{1}} \).

Ta có:  \( f(0)=\frac{1}{2}\Rightarrow f(3)=\frac{1}{f(0)}=2 \).

+ Xét  \( {{I}_{1}}=\int\limits_{0}^{3}{\frac{1}{1+f(x)}dx} \).

Đặt  \( t=3-x\Rightarrow x=3-t\Rightarrow dx=-dt  \).

Đổi cận:  \( \left\{ \begin{align} & x=0\Rightarrow t=3 \\  & x=3\Rightarrow t=0 \\ \end{align} \right. \).

Khi đó:  \( {{I}_{1}}=\int\limits_{0}^{3}{\frac{1}{1+f(x)}dx}=\int\limits_{3}^{0}{\frac{1}{1+f(3-t)}(-dt)}=\int\limits_{0}^{3}{\frac{1}{1+f(3-t)}dt}=\int\limits_{0}^{3}{\frac{1}{1+f(3-x)}dx} \)

 \( =\int\limits_{0}^{3}{\frac{1}{1+\frac{1}{f(x)}}dx}=\int\limits_{0}^{3}{\frac{f(x)}{1+f(x)}dx} \).

Suy ra:  \( 2{{I}_{1}}=\int\limits_{0}^{3}{\frac{1}{1+f(x)}dx}+\int\limits_{0}^{3}{\frac{f(x)}{1+f(x)}dx}=\int\limits_{0}^{3}{\frac{1+f(x)}{1+f(x)}dx}=\int\limits_{0}^{3}{1dx}=3\Rightarrow {{I}_{1}}=\frac{3}{2} \).

Vậy  \( I=-1+\frac{3}{2}=\frac{1}{2} \).