Biết \(\int\limits_1^2 {\frac{{{\rm{d}}x}}{{\left( {x + 1} \right)\sqrt x  + x\sqrt {x + 1} }} = \sqrt a }  - \sqrt b  - c\) v�

* Hướng dẫn giải

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

Đặt \(u = \sqrt {x + 1}  + \sqrt x \), suy ra \({\rm{d}}u = \left( {\frac{1}{{2\sqrt {x + 1} }} + \frac{1}{{2\sqrt x }}} \right)\,{\rm{d}}x \to 2{\rm{d}}u = \frac{{\sqrt x  + \sqrt {x + 1} }}{{\sqrt {x\left( {x + 1} \right)} }}\,{\rm{d}}x.\)

Đổi cận \(\left\{ \begin{array}{l}
x = 2 \to u = \sqrt 3  + \sqrt 2 \\
x = 1 \to u = \sqrt 2  + 1
\end{array} \right..\) Khi đó \(I = 2\int\limits_{\sqrt 2  + 1}^{\sqrt 3  + \sqrt 2 } {\frac{{{\rm{d}}u}}{{{u^2}}}} \, = \left. { - \frac{2}{u}} \right|_{\sqrt 2  + 1}^{\sqrt 3  + \sqrt 2 } =  - 2\left( {\frac{1}{{\sqrt 3  + \sqrt 2 }} - \frac{1}{{\sqrt 2  + 1}}} \right)\)

\( =  - 2\left( {\frac{{\sqrt 3  - \sqrt 2 }}{{3 - 2}} - \frac{{\sqrt 2  - 1}}{{2 - 1}}} \right) = \sqrt {32}  - \sqrt {12}  - 2 \to \left\{ \begin{array}{l}
a = 32\\
b = 12\\
c = 2
\end{array} \right. \to P = 46.\)