Cho \(\int\limits_{}^{} {{{dx} \over {\sqrt {x + 2} + \sqrt {x + 1} }}} = a\left( {x + 2} \right)\sqrt {x + 2} + b\left( {x + 1} \right)\sqrt {x + 1} + C\). Khi đó \(3a + b\) bằng:
A.\( - {2 \over 3}\)
B.\({1 \over 3}\)
C.\({4 \over 3}\)
D.\({2 \over 3}\)