Đáp án:
\[T = 14\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
u = \ln \left( {x + 2} \right)\\
v' = x + 1
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = \frac{1}{{x + 2}}\\
v = \frac{1}{2}{x^2} + x
\end{array} \right.\\
\int\limits_{ - 1}^2 {\left( {x + 1} \right)\ln \left( {x + 2} \right)dx} \\
= \mathop {\left( {\frac{1}{2}{x^2} + x} \right).\ln \left( {x + 2} \right)}\nolimits_{ - 1}^2 - \int\limits_{ - 1}^2 {\frac{1}{{x + 2}}.\left( {\frac{1}{2}{x^2} + x} \right)dx} \\
= \left( {\frac{{15}}{2}.\ln 4 + \frac{1}{2}.\ln 1\,\,} \right) - \frac{1}{2}\int\limits_{ - 1}^2 {\frac{{{x^2} + 2x}}{{x + 2}}dx} \\
= \left( {15\ln 2 + \frac{1}{2}} \right) - \frac{1}{2}\int\limits_{ - 1}^2 {xdx} \\
= \left( {15\ln 2 + \frac{1}{2}} \right) - \frac{1}{2}.\frac{3}{2}\\
= 15\ln 2 - \frac{1}{4}\\
\Rightarrow a = 15;\,\,\,\,b = - 1 \Rightarrow T = a + b = 14
\end{array}\)
