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Câu hỏi: \(\int {x\ln \left( {x + 1} \right)dx} \) bằng
A. \(\left( {\dfrac{{{x^2}}}{2} - 1} \right)\ln \left({x + 1} \right) + \dfrac{1}{4}{\left({x - 1} \right)^2} + C\)
B. \(\left( {\dfrac{{{x^2}}}{2} - 1} \right)\ln \left({x + 1} \right) - \dfrac{1}{2}{\left({x - 1} \right)^2} + C\)
C. \(\left( {\dfrac{{{x^2}}}{2} - \dfrac{1}{2}} \right)\ln \left({x + 1} \right) - \dfrac{1}{4}{\left({x - 1} \right)^2} + C\)
D. \(\left( {\dfrac{{{x^2}}}{2} + 1} \right)\ln \left({x + 1} \right) - \dfrac{1}{4}{\left({x - 1} \right)^2} + C\)
A. \(\left( {\dfrac{{{x^2}}}{2} - 1} \right)\ln \left({x + 1} \right) + \dfrac{1}{4}{\left({x - 1} \right)^2} + C\)
B. \(\left( {\dfrac{{{x^2}}}{2} - 1} \right)\ln \left({x + 1} \right) - \dfrac{1}{2}{\left({x - 1} \right)^2} + C\)
C. \(\left( {\dfrac{{{x^2}}}{2} - \dfrac{1}{2}} \right)\ln \left({x + 1} \right) - \dfrac{1}{4}{\left({x - 1} \right)^2} + C\)
D. \(\left( {\dfrac{{{x^2}}}{2} + 1} \right)\ln \left({x + 1} \right) - \dfrac{1}{4}{\left({x - 1} \right)^2} + C\)
Phương pháp giải
Tìm nguyên hàm bằng phương pháp từng phần \(\int {udv} = uv - \int {vdu} \).
Lời giải chi tiết
Đặt \(\left\{ \begin{array}{l}u = \ln \left( {x + 1} \right)\\dv = xdx\end{array} \right.\) \(\Rightarrow \left\{ \begin{array}{l}du = \dfrac{1}{{x + 1}}dx\\v = \dfrac{{{x^2}}}{2}\end{array} \right.\)
Khi đó \(\int {x\ln \left( {x + 1} \right)dx} \)\(= \dfrac{{{x^2}}}{2}\ln \left( {x + 1} \right) - \dfrac{1}{2}\int {\dfrac{{{x^2}}}{{x + 1}}dx} \) \(= \dfrac{{{x^2}}}{2}\ln \left( {x + 1} \right) - \dfrac{1}{2}\int {\left({x - 1 + \dfrac{1}{{x + 1}}} \right)dx} \)
\(= \dfrac{{{x^2}}}{2}\ln \left( {x + 1} \right) - \dfrac{{{x^2}}}{4} + \dfrac{1}{2}x - \dfrac{1}{2}\ln \left({x + 1} \right) + C\) \(= \left( {\dfrac{{{x^2}}}{2} - \dfrac{1}{2}} \right)\ln \left({x + 1} \right) - \dfrac{1}{4}\left({{x^2} - 2x + 1} \right) + C'\)
\(= \left( {\dfrac{{{x^2}}}{2} - \dfrac{1}{2}} \right)\ln \left({x + 1} \right) - \dfrac{1}{4}{\left({x - 1} \right)^2} + C'\)
Tìm nguyên hàm bằng phương pháp từng phần \(\int {udv} = uv - \int {vdu} \).
Lời giải chi tiết
Đặt \(\left\{ \begin{array}{l}u = \ln \left( {x + 1} \right)\\dv = xdx\end{array} \right.\) \(\Rightarrow \left\{ \begin{array}{l}du = \dfrac{1}{{x + 1}}dx\\v = \dfrac{{{x^2}}}{2}\end{array} \right.\)
Khi đó \(\int {x\ln \left( {x + 1} \right)dx} \)\(= \dfrac{{{x^2}}}{2}\ln \left( {x + 1} \right) - \dfrac{1}{2}\int {\dfrac{{{x^2}}}{{x + 1}}dx} \) \(= \dfrac{{{x^2}}}{2}\ln \left( {x + 1} \right) - \dfrac{1}{2}\int {\left({x - 1 + \dfrac{1}{{x + 1}}} \right)dx} \)
\(= \dfrac{{{x^2}}}{2}\ln \left( {x + 1} \right) - \dfrac{{{x^2}}}{4} + \dfrac{1}{2}x - \dfrac{1}{2}\ln \left({x + 1} \right) + C\) \(= \left( {\dfrac{{{x^2}}}{2} - \dfrac{1}{2}} \right)\ln \left({x + 1} \right) - \dfrac{1}{4}\left({{x^2} - 2x + 1} \right) + C'\)
\(= \left( {\dfrac{{{x^2}}}{2} - \dfrac{1}{2}} \right)\ln \left({x + 1} \right) - \dfrac{1}{4}{\left({x - 1} \right)^2} + C'\)
Đáp án C.