Biết \(I = \int\limits_1^2 {\frac{{{\rm{d}}x}}{{\l...

D

- Hướng dẫn giải

Ta có: \(\sqrt {x + 1}  - \sqrt x  \ne 0\), \(\forall x \in \left[ {1;2} \right]\) nên:

\(I = \int\limits_1^2 {\frac{{{\rm{d}}x}}{{\left( {x + 1} \right)\sqrt x  + x\sqrt {x + 1} }}}  = \int\limits_1^2 {\frac{{{\rm{d}}x}}{{\sqrt {x\left( {x + 1} \right)} \left( {\sqrt {x + 1}  + \sqrt x } \right)}}} \)

\(\begin{array}{l}
 = \int\limits_1^2 {\frac{{\left( {\sqrt {x + 1}  - \sqrt x } \right){\rm{d}}x}}{{\sqrt {x\left( {x + 1} \right)} \left( {\sqrt {x + 1}  + \sqrt x } \right)\left( {\sqrt {x + 1}  - \sqrt x } \right)}}}  = \int\limits_1^2 {\frac{{\left( {\sqrt {x + 1}  - \sqrt x } \right){\rm{d}}x}}{{\sqrt {x\left( {x + 1} \right)} }}} \\
 = \int\limits_1^2 {\left( {\frac{1}{{\sqrt x }} - \frac{1}{{\sqrt {x + 1} }}} \right){\rm{d}}x}  = \left. {\left( {2\sqrt x  - 2\sqrt {x + 1} } \right)} \right|_1^2 = 4\sqrt 2  - 2\sqrt 3  - 2 = \sqrt {32}  - \sqrt {12}  - 2
\end{array}\).

Mà \(I = \sqrt a  - \sqrt b  - c\) nên \(\left\{ \begin{array}{l}
a = 32\\
b = 12\\
c = 2
\end{array} \right.\). Suy ra: \(P = a + b + c = 32 + 12 + 2 = 46\).