Tính tích phân I = intlimits_1^{{2^{1000}}} (lnx/(x+1)^2dx)

Xét tích phân: \(I = \int\limits_1^{{2^{1000}}} {\frac{{\ln x}}{{{{\left( {x + 1} \right)}^2}}}dx}\)

Đặt: \(\left\{ \begin{array}{l} u = \ln x\\ dv = \frac{1}{{{{(x + 1)}^2}}}dx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = \frac{1}{x}dx\\ v = - \frac{1}{{x + 1}} \end{array} \right.\)

Khi đó: 

\(I = - \frac{{\ln x}}{{x + 1}}\left| \begin{array}{l} ^{{2^{1000}}}\\ _1 \end{array} \right. + \int\limits_1^{{2^{1000}}} {\frac{1}{{x + 1}}.\frac{1}{x}dx}\)

\(= - \frac{{\ln {2^{1000}}}}{{1 + {2^{1000}}}} + \int\limits_1^{{2^{1000}}} {\frac{1}{{x + 1}}.\frac{1}{x}dx} = - \frac{{1000\ln 2}}{{1 + {2^{1000}}}} + \int\limits_1^{{2^{1000}}} {\left( {\frac{1}{x} - \frac{1}{{x + 1}}} \right)dx}\)

\(= - \frac{{1000\ln 2}}{{1 + {2^{1000}}}} + \left( {\ln \left| x \right| - \ln \left| {x + 1} \right|} \right)\left| \begin{array}{l} ^{{2^{1000}}}\\ _1 \end{array} \right. = - \frac{{1000\ln 2}}{{1 + {2^{1000}}}} + \ln \left| {\frac{x}{{x + 1}}} \right|\left| \begin{array}{l} ^{{2^{1000}}}\\ _1 \end{array} \right.\)

\(= - \frac{{1000\ln 2}}{{1 + {2^{1000}}}} + \ln \frac{{{2^{1000}}}}{{1 + {2^{1000}}}}.\)