Google
Câu hỏi: Biết tích phân $I=\int\limits_{1}^{10}{\dfrac{\log x}{{{\left( x+1 \right)}^{2}}}dx=a+b}\log 2+c\log 11$, trong đó $a,b,c$ là các số hữu tỷ. Tính $S=11a+2b+3c$.
A. $11.$
B. $9.$
C. $-9.$
D. $-11.$
Đặt $\left\{ \begin{aligned}
& u=\log x \\
& dv=\dfrac{1}{{{\left( x+1 \right)}^{2}}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{1}{x\ln 10}dx \\
& v=-\dfrac{1}{x+1} \\
\end{aligned} \right.$
$\begin{aligned}
& I=\int\limits_{1}^{10}{\dfrac{\log x}{{{\left( x+1 \right)}^{2}}}dx=-\dfrac{1}{x+1}\log x\left| \begin{aligned}
& 10 \\
& 1 \\
\end{aligned} \right.+}\dfrac{1}{\ln 10}\int\limits_{1}^{10}{\dfrac{dx}{x\left( x+1 \right)}}=-\dfrac{1}{11}+\dfrac{1}{\ln 10}\int\limits_{1}^{10}{\left( \dfrac{1}{x}-\dfrac{1}{x+1} \right)dx} \\
& \text{ }=-\dfrac{1}{11}+\dfrac{1}{\ln 10}\left( \ln x-\ln \left( x+1 \right) \right)\left| \begin{aligned}
& 10 \\
& 1 \\
\end{aligned} \right.=-\dfrac{1}{11}+\dfrac{1}{\ln 10}\left( \ln 10-\ln 11+\ln 2 \right)=\dfrac{10}{11}+\log 2-\log 11 \\
\end{aligned}$
Do đó suy ra $\left\{ \begin{aligned}
& a=\dfrac{10}{11} \\
& b=1 \\
& c=-1 \\
\end{aligned} \right.\Rightarrow S=11.\dfrac{10}{11}+2.1+3.\left( -1 \right)=9$.
A. $11.$
B. $9.$
C. $-9.$
D. $-11.$
Đặt $\left\{ \begin{aligned}
& u=\log x \\
& dv=\dfrac{1}{{{\left( x+1 \right)}^{2}}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{1}{x\ln 10}dx \\
& v=-\dfrac{1}{x+1} \\
\end{aligned} \right.$
$\begin{aligned}
& I=\int\limits_{1}^{10}{\dfrac{\log x}{{{\left( x+1 \right)}^{2}}}dx=-\dfrac{1}{x+1}\log x\left| \begin{aligned}
& 10 \\
& 1 \\
\end{aligned} \right.+}\dfrac{1}{\ln 10}\int\limits_{1}^{10}{\dfrac{dx}{x\left( x+1 \right)}}=-\dfrac{1}{11}+\dfrac{1}{\ln 10}\int\limits_{1}^{10}{\left( \dfrac{1}{x}-\dfrac{1}{x+1} \right)dx} \\
& \text{ }=-\dfrac{1}{11}+\dfrac{1}{\ln 10}\left( \ln x-\ln \left( x+1 \right) \right)\left| \begin{aligned}
& 10 \\
& 1 \\
\end{aligned} \right.=-\dfrac{1}{11}+\dfrac{1}{\ln 10}\left( \ln 10-\ln 11+\ln 2 \right)=\dfrac{10}{11}+\log 2-\log 11 \\
\end{aligned}$
Do đó suy ra $\left\{ \begin{aligned}
& a=\dfrac{10}{11} \\
& b=1 \\
& c=-1 \\
\end{aligned} \right.\Rightarrow S=11.\dfrac{10}{11}+2.1+3.\left( -1 \right)=9$.
Đáp án B.