|
|
Ta có:
Đặt $u = \ln x\,\,\, \Rightarrow \,\,\,du = \frac{{dx}}{x}$
$dv = \frac{{dx}}{{{{\left( {x + 1} \right)}^2}}}\,\,\,\, \Rightarrow \,\,\,\,\,v = -
\frac{1}{{x + 1}}$
$\begin{array}{l}
I = - \left. {\frac{{\ln x}}{{x + 1}}} \right]_{\frac{1}{e}}^e + \int\limits_{\frac{1}{e}}^e
{\frac{{dx}}{{x\left( {x + 1} \right)}}} = - \left( {\frac{1}{{e + 1}} + \frac{e}{{e + 1}}} \right) +
\int\limits_{\frac{1}{e}}^e {\left( {\frac{1}{x} - \frac{1}{{x + 1}}} \right)dx} \\
\,\,\, = - 1 + \left. {\ln \left| x \right|} \right]_{\frac{1}{e}}^e - \left. {\ln \left| {x + 1} \right|}
\right]_{\frac{1}{e}}^e = - 1 + \left( {1 + 1} \right) - \ln e = 0
\end{array}$
Vậy $I = 0$
|