Tính tích phân \(I = \int\limits_1^{{2^{1000}}} {\dfrac{{\ln x}}{{{{(x + 1)}^2}}}dx} \)
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A.
\(I = - \dfrac{{\ln {2^{1000}}}}{{1 + {2^{1000}}}} + \ln \dfrac{2^{1001}}{{1 + {2^{1000}}}}\)
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B.
\(I = - \dfrac{{1000\ln 2}}{{1 + {2^{1000}}}} + \ln \dfrac{{{2^{1000}}}}{{1 + {2^{1000}}}}\)
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C.
\(I = \dfrac{{\ln {2^{1000}}}}{{1 + {2^{1000}}}} - 1001\ln \dfrac{2}{{1 + {2^{1000}}}}\)
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D.
\(I = \dfrac{{1000\ln 2}}{{1 + {2^{1000}}}} - \ln \dfrac{{{2^{1000}}}}{{1 + {2^{1000}}}}\)
- Bước 1: Đặt \(\left\{ \begin{array}{l}u = \ln \left( {ax + b} \right)\\dv = f\left( x \right)dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \dfrac{a}{{\left( {ax + b} \right)}}dx\\v = \int {f\left( x \right)dx} \end{array} \right.\)
- Bước 2: Tính tích phân theo công thức \(\int\limits_m^n {f\left( x \right)\ln \left( {ax + b} \right)dx} = \left. {uv} \right|_m^n - \int\limits_m^n {vdu} \)
Đặt $\left\{ \begin{array}{l}u = \ln x\\dv = \dfrac{{dx}}{{{{(x + 1)}^2}}}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \dfrac{{dx}}{x}\\v = - \dfrac{1}{{x + 1}}\end{array} \right.$
$\begin{array}{l} \Rightarrow I = - \dfrac{{\ln x}}{{x + 1}}\left| {_1^{{2^{1000}}} + \int\limits_1^{{2^{1000}}} {\dfrac{1}{{x + 1}}.} \dfrac{{dx}}{x} = } \right. - \dfrac{{\ln {2^{1000}}}}{{{2^{1000}} + 1}} + \int\limits_1^{{2^{1000}}} {\left( {\dfrac{1}{x} - \dfrac{1}{{x + 1}}} \right)dx} \\ = - \dfrac{{1000\ln 2}}{{{2^{1000}} + 1}} + \left. {\ln \left| {\dfrac{x}{{x + 1}}} \right|} \right|_1^{{2^{1000}}} = - \dfrac{{1000\ln 2}}{{{2^{1000}} + 1}} + \ln \dfrac{{{2^{1000}}}}{{{2^{1000}} + 1}} - \ln \dfrac{1}{2} \\ = - \dfrac{{1000\ln 2}}{{{2^{1000}} + 1}} + \ln \dfrac{{{2^{1001}}}}{{{2^{1000}} + 1}}\end{array}$
Đáp án : A
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