Cho hàm số \(f\left( x \right)\) có \(f\left( 3 \right) = - \frac{3}\) và \(f'\left( x \right) = \frac{x}{{\sqrt {x + 1} - 1}}\). Khi đó \(\int\limits_3^8 {f\left( x \right)dx} \) bằng

Đáp án C

Ta có: \(f'\left( x \right) = \frac{x}{{\sqrt {x + 1} - 1}} = \frac{{x\left( {\sqrt {x + 1} + 1} \right)}}{{\left( {\sqrt {x + 1} - 1} \right)\left( {\sqrt {x + 1} + 1} \right)}} = \sqrt {x + 1} + 1\)

\[ \Rightarrow f\left( x \right) = \int {\left( {\sqrt {x + 1} + 1} \right)dx} = \frac{2}{3}\sqrt {{{\left( {x + 1} \right)}^3}} + x + C\]

Do \[f\left( 3 \right) = - \frac{3} \Rightarrow \frac{2}{3}\left( {\sqrt {{{\left( {3 + 1} \right)}^3}} } \right) + 3 + C = - \frac{3} \Leftrightarrow C = - \frac{3}\].

Từ đó: \[f\left( x \right) = \frac{2}{3}\sqrt {{{\left( {x + 1} \right)}^3}} + x - \frac{3}\]

\[ \Rightarrow \int\limits_8^8 {f\left( x \right)dx} = \int\limits_3^8 {\left[ {\frac{2}{3}\sqrt {{{\left( {x + 1} \right)}^3}} + x - \frac{3}} \right]dx} = \left. {\left( {\frac{4}\sqrt[3]{{{{\left( {x + 1} \right)}^5}}} + \frac{{{x^2}}}{2} - \frac{3}} \right)} \right|_3^8 = \frac\].

Vậy \[\int\limits_3^8 {f\left( x \right)dx} = \frac\].