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Câu hỏi: Cho $\int\limits_{-1}^{4}{x\ln \left( x+2 \right)dx=a\ln 6+\dfrac{5}{b}}$ với a, b là các số nguyên dương. Giá trị $2a+3b$ bằng:
A. 24.
B. 26.
C. 27.
D. 23.
A. 24.
B. 26.
C. 27.
D. 23.
Đặt $\left\{ \begin{aligned}
& \ln \left( x+2 \right)=u \\
& xdx=dv \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \dfrac{1}{x+2}dx=du \\
& \dfrac{{{x}^{2}}}{2}=v \\
\end{aligned} \right.$
Suy ra $\int\limits_{-1}^{4}{x\ln \left( x+2 \right)dx=\left. \ln \left( x+2 \right).\dfrac{{{x}^{2}}}{2} \right|_{-1}^{4}-\int\limits_{-1}^{4}{\dfrac{{{x}^{2}}}{2}.\dfrac{1}{x+2}dx}}$
$=8\ln 6-\dfrac{1}{2}\int\limits_{-1}^{4}{\left( x-2+\dfrac{4}{x+2} \right)dx=8\ln 6-\dfrac{1}{2}\left. \left( \dfrac{{{x}^{2}}}{2}-2x+4\ln \left| x \right|+2 \right) \right|_{-1}^{4}}$
$=8\ln 6-\dfrac{1}{2}\left( 4\ln 6-\dfrac{5}{2} \right)=6\ln 6+\dfrac{5}{4}$
Theo giả thiết $\int\limits_{-1}^{4}{x\ln \left( x+2 \right)dx=a\ln 6+\dfrac{5}{b}}$ nên suy ra $a=6;b=4\Rightarrow 2a+3b=2.6+3.4=24$.
& \ln \left( x+2 \right)=u \\
& xdx=dv \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \dfrac{1}{x+2}dx=du \\
& \dfrac{{{x}^{2}}}{2}=v \\
\end{aligned} \right.$
Suy ra $\int\limits_{-1}^{4}{x\ln \left( x+2 \right)dx=\left. \ln \left( x+2 \right).\dfrac{{{x}^{2}}}{2} \right|_{-1}^{4}-\int\limits_{-1}^{4}{\dfrac{{{x}^{2}}}{2}.\dfrac{1}{x+2}dx}}$
$=8\ln 6-\dfrac{1}{2}\int\limits_{-1}^{4}{\left( x-2+\dfrac{4}{x+2} \right)dx=8\ln 6-\dfrac{1}{2}\left. \left( \dfrac{{{x}^{2}}}{2}-2x+4\ln \left| x \right|+2 \right) \right|_{-1}^{4}}$
$=8\ln 6-\dfrac{1}{2}\left( 4\ln 6-\dfrac{5}{2} \right)=6\ln 6+\dfrac{5}{4}$
Theo giả thiết $\int\limits_{-1}^{4}{x\ln \left( x+2 \right)dx=a\ln 6+\dfrac{5}{b}}$ nên suy ra $a=6;b=4\Rightarrow 2a+3b=2.6+3.4=24$.
Đáp án A.