Cho \(\int\limits_0^1 {\frac{{d{\rm{x}}}}{{\sqrt {x + 2}  + \sqrt {x + 1} }}}  = a\sqrt b  - \frac{8}{3}\sqrt a  + \frac{2}{3}\lef

Câu hỏi :

Cho \(\int\limits_0^1 {\frac{{d{\rm{x}}}}{{\sqrt {x + 2}  + \sqrt {x + 1} }}}  = a\sqrt b  - \frac{8}{3}\sqrt a  + \frac{2}{3}\left( {a,b \in {R^*}} \right).\) Tính \(a + 2b\)?