Tìm nguyên hàm logx /x đx

Đáp án:

\[\int {\frac{{\log x}}{x}dx}  = \frac{{{{\left( {\ln x} \right)}^2}}}{{2\ln 10}}\]

Giải thích các bước giải:

\(\begin{array}{l}
\int {\frac{{\log x}}{x}dx} \\
\left\{ \begin{array}{l}
u = \log x\\
v' = \frac{1}{x}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = \frac{1}{{x.\ln 10}}\\
v = \ln x
\end{array} \right.\\
 \Rightarrow I = \int {\frac{{\log x}}{x}dx} \\
 = \log x.\ln x - \int {\frac{{\ln x}}{{x.\ln 10}}dx} \\
 = \log e.{\log _e}x.\ln x - \frac{1}{{\ln 10}}\int {\frac{{\ln x}}{x}dx} \\
 = \log e.{\left( {\ln x} \right)^2} - \frac{1}{{\ln 10}}.\int {\frac{{\ln x}}{x}dx} \\
\left\{ \begin{array}{l}
u = \ln x\\
v' = \frac{1}{x}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = \frac{1}{x}\\
v = \ln x
\end{array} \right.\\
 \Rightarrow \int {\frac{{\ln x}}{x}dx}  = {\left( {\ln x} \right)^2} - \int {\frac{{\ln x}}{x}dx}  \Rightarrow \int {\frac{{\ln x}}{x}dx}  = \frac{{{{\left( {\ln x} \right)}^2}}}{2}\\
 \Rightarrow I = \log e.{\left( {\ln x} \right)^2} - \frac{1}{{\ln 10}}.\frac{{{{\left( {\ln x} \right)}^2}}}{2} = {\left( {\ln x} \right)^2}.\left( {\log e - \frac{1}{{2\ln 10}}} \right)\\
 = {\left( {\ln x} \right)^2}.\left( {\frac{1}{{\ln 10}} - \frac{1}{{2\ln 10}}} \right)\\
 = \frac{{{{\left( {\ln x} \right)}^2}}}{{2\ln 10}}
\end{array}\)