Cho hàm số f(x) có f(3)=-25/3 và f'(x)=x/(căn bậc hai của (x+1))-1 . Khi đó

Đáp án C

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

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

Do \(f\left( 3 \right) = - \frac{{25}}{3} \Rightarrow \frac{2}{3}\sqrt {{{\left( {3 + 1} \right)}^3}} + 3 + C = - \frac{{25}}{3} \Leftrightarrow C = - \frac{{50}}{3}\).

Từ đó: \(f\left( x \right) = \frac{2}{3}\sqrt {{{\left( {x + 1} \right)}^3}} + x - \frac{{50}}{3}\).

\[ \Rightarrow \int\limits_3^8 {f\left( x \right)dx} = \int\limits_3^8 {\left[ {\frac{2}{3}\sqrt {{{\left( {x + 1} \right)}^3}} + x - \frac{{50}}{3}} \right]dx} = \left. {\left( {\frac{4}{{15}}\sqrt[3]{{{{\left( {x + 1} \right)}^5}}} + \frac{{{x^2}}}{2} - \frac{{50}}{3}x} \right)} \right|_3^8 = \frac{{13}}{{30}}\].

Vậy \(\int\limits_3^8 {f\left( x \right)d{\rm{x}}} = \frac{{13}}{{30}}\).