Lời giải chi tiết:

Do \(y = f\left( x \right)\) liên tục, nhận giá trị dương trên \(\left( {0; + \infty } \right)\)  nên \({\left[ {f'\left( x \right)} \right]^2} = \left( {x + 1} \right)f\left( x \right)\)

\(\begin{array}{l} \Rightarrow f'\left( x \right) = \sqrt {x + 1} .\sqrt {f\left( x \right)} ,\,\,\forall x \in \left( {0; + \infty } \right) \Rightarrow \dfrac{{f'\left( x \right)}}{{\sqrt {f\left( x \right)} }} = \sqrt {x + 1}  \Rightarrow \int\limits_3^8 {\dfrac{{f'\left( x \right)}}{{\sqrt {f\left( x \right)} }}dx}  = \int\limits_3^8 {\sqrt {x + 1} dx} \\ \Leftrightarrow \int\limits_3^8 {\dfrac{{d\left( {f\left( x \right)} \right)}}{{\sqrt {f\left( x \right)} }}}  = \left. {\dfrac{2}{3}\left( {x + 1} \right)\sqrt {x + 1} } \right|_3^8 \Leftrightarrow \left. {2\sqrt {f\left( x \right)} } \right|_3^8 = \dfrac{2}{3}.\left( {27 - 8} \right) \Leftrightarrow 2\left( {\sqrt {f\left( 8 \right)}  - \sqrt {f\left( 3 \right)} } \right) = 2.\dfrac{{19}}{3}\\ \Leftrightarrow \sqrt {f\left( 8 \right)}  - \sqrt {\dfrac{2}{3}}  = \dfrac{{19}}{3} \Rightarrow {f^2}\left( 8 \right) \approx 2613,3 \Rightarrow 2613 < {f^2}\left( 8 \right) < 2614\end{array}\)

Chọn: C