Cho \(\int\limits_0^1 {{{\left( {\frac{{2x + 1}}{{...

C

- Hướng dẫn giải

\(\begin{array}{l}
\int\limits_0^1 {{{\left( {\frac{{2x + 1}}{{x + 1}}} \right)}^2}dx}  = \int\limits_0^1 {{{\left( {2 - \frac{1}{{x + 1}}} \right)}^2}}  = \int\limits_0^1 {\left( {4 - \frac{4}{{x + 1}} + \frac{1}{{{{\left( {x + 1} \right)}^2}}}} \right)dx} \\
 = \left. {\left( {4 - 4\ln \left| {x + 1} \right| - \frac{1}{{x + 1}}} \right)} \right|\begin{array}{*{20}{c}}
1\\
0
\end{array} = \frac{9}{2} - 4\ln 2 \Rightarrow a = \frac{9}{2},b =  - 4 \Rightarrow P = 5
\end{array}\)