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Câu hỏi: Biết $\int\limits_{0}^{2}{2x\ln \left( x+1 \right)\text{d}x=a\ln b}$, với $a,b\in {{\mathbb{N}}^{*}}$. Tính $T=a+b$.
A. $T=6$.
B. $T=8$.
C. $T=7$.
D. $T=5$.
A. $T=6$.
B. $T=8$.
C. $T=7$.
D. $T=5$.
Đặt: $\left\{ \begin{matrix}
u=\ln \left( x+1 \right) \\
\text{d}v=2x\text{d}x \\
\end{matrix} \right.\Rightarrow \left\{ \begin{matrix}
\text{d}u=\dfrac{\text{d}x}{x+1} \\
v={{x}^{2}} \\
\end{matrix} \right.$
$\int\limits_{0}^{2}{2x\ln \left( x+1 \right)\text{d}x=}\left. {{x}^{2}}\ln \left( x+1 \right) \right|_{0}^{2}-\int\limits_{0}^{2}{\dfrac{{{x}^{2}}\text{d}x}{x+1}}=\left. {{x}^{2}}\ln \left( x+1 \right) \right|_{0}^{2}-\int\limits_{0}^{2}{\left( x-1 \right)\text{d}x}-\int\limits_{0}^{2}{\dfrac{\text{d}x}{x+1}}$
$=4\ln 3-\left. \left( \dfrac{{{x}^{2}}}{2}-x \right) \right|_{0}^{2}-\left. \ln \left( x+1 \right) \right|_{0}^{2}=3\ln 3$
$\Rightarrow \left\{ \begin{matrix}
a=3 \\
b=3 \\
\end{matrix}\Rightarrow T=a+b=6 \right.$
u=\ln \left( x+1 \right) \\
\text{d}v=2x\text{d}x \\
\end{matrix} \right.\Rightarrow \left\{ \begin{matrix}
\text{d}u=\dfrac{\text{d}x}{x+1} \\
v={{x}^{2}} \\
\end{matrix} \right.$
$\int\limits_{0}^{2}{2x\ln \left( x+1 \right)\text{d}x=}\left. {{x}^{2}}\ln \left( x+1 \right) \right|_{0}^{2}-\int\limits_{0}^{2}{\dfrac{{{x}^{2}}\text{d}x}{x+1}}=\left. {{x}^{2}}\ln \left( x+1 \right) \right|_{0}^{2}-\int\limits_{0}^{2}{\left( x-1 \right)\text{d}x}-\int\limits_{0}^{2}{\dfrac{\text{d}x}{x+1}}$
$=4\ln 3-\left. \left( \dfrac{{{x}^{2}}}{2}-x \right) \right|_{0}^{2}-\left. \ln \left( x+1 \right) \right|_{0}^{2}=3\ln 3$
$\Rightarrow \left\{ \begin{matrix}
a=3 \\
b=3 \\
\end{matrix}\Rightarrow T=a+b=6 \right.$
Đáp án A.