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

B

- Hướng dẫn giải

Ta có: \(I = \int\limits_0^2 {2x\ln \left( {1 + x} \right)dx} \) 

Đặt \(\left\{ \begin{array}{l}
u = \ln \left( {1 + x} \right)\\
dv = 2xdx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = \frac{1}{{x + 1}}dx\\
v = {x^2}
\end{array} \right.\) 

\(\begin{array}{l}
 \Rightarrow I = {x^2}.\ln \left( {x + 1} \right)\left| \begin{array}{l}
^2\\
_0
\end{array} \right. - \int\limits_0^2 {\frac{{{x^2}}}{{x + 1}}dx}  = 4\ln 3 - \int\limits_0^2 {\left( {x - 1 + \frac{1}{{x + 1}}} \right)} dx\\
 = 4\ln 3 - \left( {\frac{{{x^2}}}{2} - x + \ln \left| {x + 1} \right|} \right)\left| \begin{array}{l}
^2\\
_0
\end{array} \right. = 4\ln 3 - \left( {0 + \ln 3 - 0} \right) = 3\ln 3\\
 \Rightarrow \left\{ \begin{array}{l}
a = 3\\
b = 3
\end{array} \right. \Rightarrow 3a + 4b = 3.3 + 4.3 = 21
\end{array}\)