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Câu hỏi: Biêt $\int\limits_{1}^{5}{\dfrac{\ln x}{{{x}^{2}}}}dx=a.\ln 5+b$ với a, b là các số hữu tỉ. Tính tích a.b
A. $ab=-\dfrac{4}{25}$
B. $ab=\dfrac{4}{25}$
C. $ab=-\dfrac{6}{25}$
D. $ab=\dfrac{6}{25}$
A. $ab=-\dfrac{4}{25}$
B. $ab=\dfrac{4}{25}$
C. $ab=-\dfrac{6}{25}$
D. $ab=\dfrac{6}{25}$
Đặt $\left\{ \begin{aligned}
& u=\ln x \\
& dv=\dfrac{1}{{{x}^{2}}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{1}{x}dx \\
& v=-\dfrac{1}{x} \\
\end{aligned} \right.$
$\int\limits_{1}^{5}{\dfrac{\ln x}{{{x}^{2}}}}dx=\left. -\dfrac{1}{x}\ln x \right|_{1}^{5}+\int\limits_{1}^{5}{\dfrac{1}{{{x}^{2}}}}dx=-\dfrac{1}{x}\ln 5-\left. \dfrac{1}{x} \right|_{1}^{5}=-\dfrac{1}{x}\ln 5+\dfrac{4}{5}$
Suy ra $ab=-\dfrac{4}{25}$
& u=\ln x \\
& dv=\dfrac{1}{{{x}^{2}}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{1}{x}dx \\
& v=-\dfrac{1}{x} \\
\end{aligned} \right.$
$\int\limits_{1}^{5}{\dfrac{\ln x}{{{x}^{2}}}}dx=\left. -\dfrac{1}{x}\ln x \right|_{1}^{5}+\int\limits_{1}^{5}{\dfrac{1}{{{x}^{2}}}}dx=-\dfrac{1}{x}\ln 5-\left. \dfrac{1}{x} \right|_{1}^{5}=-\dfrac{1}{x}\ln 5+\dfrac{4}{5}$
Suy ra $ab=-\dfrac{4}{25}$
Đáp án A.