Tính tích phân I = tích phân từ 1 đến 2^1000 ln x / ( x + 1 )^2 d x

Đặt\(\left\{ {\begin{array}{*{20}{c}}{u = \ln x}\\{dv = \frac{{dx}}{{{{(x + 1)}^2}}}}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = \frac{{dx}}{x}}\\{v = - \frac{1}{{x + 1}}}\end{array}} \right.\)

\(\begin{array}{l} \Rightarrow I = - \frac{{lnx}}{{x + 1}}\left| {_1^{{2^{1000}}}} \right. + \int\limits_1^{{2^{1000}}} {\frac{1}{{x + 1}}} .\frac{{dx}}{x}\\ = - \frac{{\ln {2^{1000}}}}{{{2^{1000}} + 1}} + \int\limits_1^{{2^{1000}}} {\left( {\frac{1}{x} - \frac{1}{{x + 1}}} \right)} dx\\ = - \frac{{1000ln2}}{{{2^{1000}} + 1}} + \ln \left| {\frac{x}{{x + 1}}} \right|\left| {_1^{{2^{1000}}}} \right.\\ = - \frac{{1000ln2}}{{{2^{1000}} + 1}} + \ln \frac{{{2^{1000}}}}{{{2^{1000}} + 1}} - \ln \frac{1}{2}\\ = - \frac{{1000ln2}}{{{2^{1000}} + 1}} + \ln \frac{{{2^{1001}}}}{{{2^{1000}} + 1}}\end{array}\)

Đáp án cần chọn là: A