Cho hàm số \(f\left( x \right)\) thỏa mãn \(f'\left( x \right) = a{x^2} + \frac{b}{{{x^3}}},\,\,f'\left( 1 \right) = 3,\,\,f\left( 1 \right) = 2,\,\,f\left( {\frac{1}{2}} \right) = - \frac{1}.\) Khi đó \(2a + b\) bằng

Đặt \(t = \sqrt {x + 1}  > 1 \Rightarrow {t^2} = x + 1 \Rightarrow x = {t^2} - 1\)

\[ \Rightarrow f'\left( x \right) = \frac{{{t^2} - 1}}{{{t^2} - t}} = \frac{{\left( {t - 1} \right)\left( {t + 1} \right)}}{{\left( {t - 1} \right)t}} = \frac{t} = 1 + \frac{1}{t} = \frac{1}{{\sqrt {x + 1} }} + 1\]

\( \Rightarrow f\left( x \right) = \int {f'\left( x \right)} \,\,dx = \int {\left( {1 + \frac{1}{{\sqrt {x + 1} }}} \right)} \,\,{\rm{d}}x = x + \int {\frac{{{\rm{d}}\left( {x + 1} \right)}}{{\sqrt {x + 1} }}} \)\[ = x + 2\int {\frac{{{\rm{d}}\left( {x + 1} \right)}}{{2\sqrt {x + 1} }}}  = x + 2\sqrt {x + 1}  + C\].

Mà \(f\left( 3 \right) = 3 \Rightarrow 3 + 2 \cdot \sqrt {3 + 1}  + C = 3 \Rightarrow C =  - 4 \Rightarrow f\left( x \right) = x + 2\sqrt {x + 1}  - 4\).

Suy ra \[\int\limits_3^8 {f\left( x \right)} \,{\rm{d}}x = \int\limits_3^8 {\left( {x + 2\sqrt {x + 1}  - 4} \right)} \,{\rm{d}}x = \frac{6}\]. Chọn B.