Cho hàm số f(x) có f'(x)=(x+4)*căn bậc hai của(x-1) với mọi x>-1

Đáp án B

Xét \(I = \int {\left( {x + 4} \right)\sqrt {x + 1} dx} ,\) đặt \(t = \sqrt {x + 1} \Rightarrow {t^2} = x + 1 \Rightarrow 2tdt = dx.\) Khi đó:

\(I = \int {\left( {{t^2} + 3} \right)t2tdx} = \int {\left( {2{t^4} + 6{t^2}} \right)} dx = \frac{{2{t^5}}}{5} + 2{t^3} + C = \frac{{2{{\left( {x + 1} \right)}^{\frac{5}{2}}}}}{5} + 2{\left( {x + 1} \right)^{\frac{3}{2}}} + C\)

Suy ra \(f\left( x \right) = \frac{{2{{\left( {x + 1} \right)}^{\frac{5}{2}}}}}{5} + 2{\left( {x + 1} \right)^{\frac{3}{2}}} + C.\) Thay \(x = 0:\)

\(f\left( 0 \right) = \frac{2}{5} + 2 + C \Rightarrow C = - \frac{2}{5}.\) Do đó \(f\left( x \right) = \frac{2}{5}{\left( {x + } \right)^{\frac{5}{2}}} + 2{\left( {x + 1} \right)^{\frac{3}{2}}} - \frac{2}{5}.\)

Khi đó

\(\int_0^3 {\left( {\frac{2}{5}{{\left( {x + 1} \right)}^{\frac{5}{2}}} + 2{{\left( {x + 1} \right)}^{\frac{3}{2}}} - \frac{2}{5}} \right)} dx = \left( {\frac{4}{{35}}{{\left( {x + 1} \right)}^{\frac{7}{2}}} + \frac{4}{5}{{\left( {x + 1} \right)}^{\frac{5}{2}}} - \frac{2}{5}x} \right)\left| {_{\scriptstyle\atop\scriptstyle0}^{\scriptstyle3\atop\scriptstyle}} \right. = \)