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Câu hỏi: Phát biểu nào sau đây đúng
A. $\int\limits_{1}^{2}{\ln x\text{d}x=\left. x.\ln x \right|_{1}^{2}}+\int\limits_{1}^{2}{1\text{d}x.}$
B. $\int\limits_{1}^{2}{\ln x\text{d}x=\left. x.\ln x \right|_{1}^{2}}-\int\limits_{1}^{2}{1\text{d}x.}$
C. $\int\limits_{1}^{2}{\ln x\text{d}x=}x.\ln x-\int\limits_{1}^{2}{1\text{d}x.}$
D. $\int\limits_{1}^{2}{\ln x\text{d}x=}x.\ln x+\int\limits_{1}^{2}{1\text{d}x.}$
A. $\int\limits_{1}^{2}{\ln x\text{d}x=\left. x.\ln x \right|_{1}^{2}}+\int\limits_{1}^{2}{1\text{d}x.}$
B. $\int\limits_{1}^{2}{\ln x\text{d}x=\left. x.\ln x \right|_{1}^{2}}-\int\limits_{1}^{2}{1\text{d}x.}$
C. $\int\limits_{1}^{2}{\ln x\text{d}x=}x.\ln x-\int\limits_{1}^{2}{1\text{d}x.}$
D. $\int\limits_{1}^{2}{\ln x\text{d}x=}x.\ln x+\int\limits_{1}^{2}{1\text{d}x.}$
Xét $I=\int\limits_{1}^{2}{\ln x\text{d}x}$. Đặt $\left\{ \begin{aligned}
& u=\ln x \\
& \text{d}v=\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\dfrac{1}{x}\text{d}x \\
& v=x \\
\end{aligned} \right..$
Khi đó: $I=x.\left. \ln x \right|_{1}^{2}-\int\limits_{1}^{2}{1\text{d}x}.$
& u=\ln x \\
& \text{d}v=\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\dfrac{1}{x}\text{d}x \\
& v=x \\
\end{aligned} \right..$
Khi đó: $I=x.\left. \ln x \right|_{1}^{2}-\int\limits_{1}^{2}{1\text{d}x}.$
Đáp án B.