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Câu hỏi: Chọn khẳng định sai.
A. $\int{{{x}^{2}}}\ln \left( x-2 \right)\text{d}x=\dfrac{{{x}^{3}}}{3}\ln \left( x-2 \right)+\dfrac{1}{3}\int{\dfrac{{{x}^{3}}}{2-x}}\text{d}x,\forall x\in \left( 2;+\infty \right)$.
B. $\int{{{x}^{2}}}\ln \left( x-2 \right)\text{d}x=\dfrac{{{x}^{3}}}{3}\ln \left( x-2 \right)-\dfrac{1}{3}\int{\dfrac{{{x}^{2}}}{x-2}}\text{d}x,\forall x\in \left( 2;+\infty \right)$.
C. $\int{{{x}^{2}}}\ln \left( x-2 \right)\text{d}x=\dfrac{{{x}^{3}}-8}{3}\ln \left( x-2 \right)-\int{\dfrac{{{x}^{2}}+2x+4}{3}}\text{d}x,\forall x\in \left( 2;+\infty \right)$.
D. $\int{{{x}^{2}}}\ln \left( x-2 \right)\text{d}x=\dfrac{{{x}^{3}}}{3}\ln \left( x-2 \right)-\dfrac{1}{3}\int{\dfrac{{{x}^{3}}}{x-2}}\text{d}x,\forall x\in \left( 2;+\infty \right)$.
A. $\int{{{x}^{2}}}\ln \left( x-2 \right)\text{d}x=\dfrac{{{x}^{3}}}{3}\ln \left( x-2 \right)+\dfrac{1}{3}\int{\dfrac{{{x}^{3}}}{2-x}}\text{d}x,\forall x\in \left( 2;+\infty \right)$.
B. $\int{{{x}^{2}}}\ln \left( x-2 \right)\text{d}x=\dfrac{{{x}^{3}}}{3}\ln \left( x-2 \right)-\dfrac{1}{3}\int{\dfrac{{{x}^{2}}}{x-2}}\text{d}x,\forall x\in \left( 2;+\infty \right)$.
C. $\int{{{x}^{2}}}\ln \left( x-2 \right)\text{d}x=\dfrac{{{x}^{3}}-8}{3}\ln \left( x-2 \right)-\int{\dfrac{{{x}^{2}}+2x+4}{3}}\text{d}x,\forall x\in \left( 2;+\infty \right)$.
D. $\int{{{x}^{2}}}\ln \left( x-2 \right)\text{d}x=\dfrac{{{x}^{3}}}{3}\ln \left( x-2 \right)-\dfrac{1}{3}\int{\dfrac{{{x}^{3}}}{x-2}}\text{d}x,\forall x\in \left( 2;+\infty \right)$.
Đặt $\left\{ \begin{aligned}
& u=\ln \left( x-2 \right) \\
& \text{d}v={{x}^{2}}\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\dfrac{1}{x-2}\text{d}x \\
& v=\dfrac{{{x}^{3}}}{3} \\
\end{aligned} \right.$.
Suy ra $\int{{{x}^{2}}}\ln \left( x-2 \right)\text{d}x=\dfrac{{{x}^{3}}}{3}\ln \left( x-2 \right)-\dfrac{1}{3}\int{\dfrac{{{x}^{3}}}{x-2}}\text{d}x,\forall x\in \left( 2;+\infty \right)$.
& u=\ln \left( x-2 \right) \\
& \text{d}v={{x}^{2}}\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\dfrac{1}{x-2}\text{d}x \\
& v=\dfrac{{{x}^{3}}}{3} \\
\end{aligned} \right.$.
Suy ra $\int{{{x}^{2}}}\ln \left( x-2 \right)\text{d}x=\dfrac{{{x}^{3}}}{3}\ln \left( x-2 \right)-\dfrac{1}{3}\int{\dfrac{{{x}^{3}}}{x-2}}\text{d}x,\forall x\in \left( 2;+\infty \right)$.
Đáp án B.