Cho \(I = \int\limits_0^3 {\frac{{xdx}}{{\sqrt {2x + 1}  + \sqrt {x + 1} }} = \frac{1}{3}\left( {7a - b} \right)} \). Khi đó a + b bằng

* Hướng dẫn giải

\(\begin{array}{l}
I = \int\limits_0^3 {\left( {\sqrt {2x + 1}  - \sqrt {x + 1} } \right)dx = \int\limits_0^3 {\sqrt {2x + 1} } dx - \int\limits_0^3 {\sqrt {x + 1} dx} } \\
 = \left( {\frac{1}{3}{{\left( {2x + 1} \right)}^{\frac{3}{2}}} - \frac{2}{3}{{\left( {x + 1} \right)}^{\frac{3}{2}}}} \right)\left| {_0^3} \right. = \frac{1}{3}\left( {7\sqrt 7  - 15} \right)\\
 \Rightarrow \left\{ \begin{array}{l}
a = \sqrt 7 \\
b = 15
\end{array} \right. \Rightarrow a + b = 15 + \sqrt 7 
\end{array}\)