Cho hàm số \(f(x)\) xác định và có đạo hàm \(f'(x)...

D

- Hướng dẫn giải

Ta có: \(f'\left( x \right){\left( {1 + f\left( x \right)} \right)^2} = {\left[ {{{\left( {f\left( x \right)} \right)}^2}\left( {x - 1} \right)} \right]^2} \Leftrightarrow \frac{{f'\left( x \right){{\left( {1 + f\left( x \right)} \right)}^2}}}{{{{\left( {f\left( x \right)} \right)}^4}}} = {\left( {x - 1} \right)^2},\forall x \in \left[ {1;3} \right]\) 

\(\begin{array}{l}
 \Rightarrow \int\limits_1^x {\frac{{f'\left( x \right){{\left( {1 + f\left( x \right)} \right)}^2}}}{{{{\left( {f\left( x \right)} \right)}^4}}}dx = \int\limits_1^x {{{\left( {x - 1} \right)}^2}dx,\forall x \in \left[ {1;3} \right]} } \\
 \Leftrightarrow \int\limits_1^x {\left[ {\frac{1}{{{{\left( {f\left( x \right)} \right)}^4}}} + \frac{2}{{{{\left( {f\left( x \right)} \right)}^3}}} + \frac{1}{{{{\left( {f\left( x \right)} \right)}^2}}}} \right]} d\left( {f\left( x \right)} \right) = \frac{{{{\left( {x - 1} \right)}^3}}}{3}\left| \begin{array}{l}
^x\\
_1
\end{array} \right.\\
 \Leftrightarrow \left[ { - \frac{1}{{3{{\left( {f\left( x \right)} \right)}^3}}} - \frac{2}{{2{{\left( {f\left( x \right)} \right)}^2}}} - \frac{1}{{f\left( x \right)}}} \right]\left| \begin{array}{l}
^x\\
_1
\end{array} \right. = \frac{{{{\left( {x - 1} \right)}^3}}}{3} - \frac{0}{3}\\
 \Leftrightarrow \left[ { - \frac{1}{{3{{\left( {f\left( x \right)} \right)}^3}}} - \frac{2}{{2{{\left( {f\left( x \right)} \right)}^2}}} - \frac{1}{{f\left( x \right)}}} \right] - \left[ { - \frac{1}{{3{{\left( {f\left( 1 \right)} \right)}^3}}} - \frac{2}{{2{{\left( {f\left( 1 \right)} \right)}^2}}} - \frac{1}{{f\left( 1 \right)}}} \right] = \frac{{{{\left( {x - 1} \right)}^3}}}{3}\\
 \Leftrightarrow \left[ { - \frac{1}{{3{{\left( {f\left( x \right)} \right)}^3}}} - \frac{2}{{2{{\left( {f\left( x \right)} \right)}^2}}} - \frac{1}{{f\left( x \right)}}} \right] - \left[ {\frac{1}{3} - 1 + 1} \right] = \frac{{{{\left( {x - 1} \right)}^3}}}{3}\\
 \Leftrightarrow  - \frac{1}{{3{{\left( {f\left( x \right)} \right)}^3}}} - \frac{2}{{2{{\left( {f\left( x \right)} \right)}^2}}} - \frac{1}{{f\left( x \right)}} = \frac{{{{\left( {x - 1} \right)}^3} + 1}}{3}\\
 \Leftrightarrow \frac{1}{3}{\left( { - \frac{1}{{f\left( x \right)}}} \right)^3} - {\left( { - \frac{1}{{f\left( x \right)}}} \right)^2} + \left( { - \frac{1}{{f\left( x \right)}}} \right) = \frac{1}{3}{x^3} - {x^2} + x\,(*)
\end{array}\) 

Xét hàm số \(g\left( t \right) = \frac{1}{3}{t^3} - {t^2} + t\) có \(g'\left( t \right) = {t^2} - 2t + 1 \ge 0,\forall t \Rightarrow \) Hàm số đồng biến trên  

Khi đó, \((*) \Leftrightarrow g\left( { - \frac{1}{{f\left( x \right)}}} \right) = g\left( x \right) \Leftrightarrow  - \frac{1}{{f\left( x \right)}} = x \Leftrightarrow f\left( x \right) =  - \frac{1}{x}\) 

\( \Rightarrow \int\limits_1^3 {f\left( x \right)dx}  = \int\limits_1^3 {\frac{{ - 1}}{x}dx =  - \ln \left| x \right|} \left| \begin{array}{l}
^3\\
_1
\end{array} \right. =  - \ln 3 = a\ln 3 + b\left( {a,b \in } \right) \Rightarrow a =  - 1,b = 0 \Rightarrow S = a + {b^2} =  - 1\)