Phương pháp giải:

Tích phân hai vế.

Lời giải chi tiết:

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

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

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

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

\( \Rightarrow \int\limits_1^3 {f\left( x \right)dx}  = \int\limits_1^3 {\dfrac{{ - 1}}{x}dx}  =  - \left. {\ln \left| x \right|} \right|_1^3 =  - \ln 3\)\( = a\ln 3 + b\,\,\left( {a,b \in \mathbb{Z}} \right) \Rightarrow a =  - 1,\,\,b = 0\)\( \Rightarrow S = a + {b^2} =  - 1\).

Chọn: D