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Câu hỏi: Xét $u=\ln (x+1)$ và $v=x^2$, khi đó $\int\limits_{0}^{1}{udv}$ bằng
A. $\left. {{x}^{2}}\ln \left( x+1 \right) \right|_{0}^{1}-\int\limits_{0}^{1}{2x\ln \left( x+1 \right)}dx$
B. $\left. {{x}^{2}}\ln \left( x+1 \right) \right|_{0}^{1}-\int\limits_{0}^{1}{\dfrac{{{x}^{2}}}{x+1}}dx$.
C. $\left. {{x}^{2}}\ln \left( x+1 \right) \right|_{0}^{1}+\int\limits_{0}^{1}{2x\ln \left( x+1 \right)}dx$.
D. $\left. {{x}^{2}}\ln \left( x+1 \right) \right|_{0}^{1}+\int\limits_{0}^{1}{\dfrac{{{x}^{2}}}{x+1}}dx$.
A. $\left. {{x}^{2}}\ln \left( x+1 \right) \right|_{0}^{1}-\int\limits_{0}^{1}{2x\ln \left( x+1 \right)}dx$
B. $\left. {{x}^{2}}\ln \left( x+1 \right) \right|_{0}^{1}-\int\limits_{0}^{1}{\dfrac{{{x}^{2}}}{x+1}}dx$.
C. $\left. {{x}^{2}}\ln \left( x+1 \right) \right|_{0}^{1}+\int\limits_{0}^{1}{2x\ln \left( x+1 \right)}dx$.
D. $\left. {{x}^{2}}\ln \left( x+1 \right) \right|_{0}^{1}+\int\limits_{0}^{1}{\dfrac{{{x}^{2}}}{x+1}}dx$.
Ta có: $\int\limits_{0}^{1}{udv}=\left. uv \right|_{0}^{1}-\int\limits_{0}^{1}{vdu}=\left. {{x}^{2}}\ln \left( x+1 \right) \right|_{0}^{1}-\int\limits_{0}^{1}{{{x}^{2}}}d\ln \left( x+1 \right)=\left. {{x}^{2}}\ln \left( x+1 \right) \right|_{0}^{1}-\int\limits_{0}^{1}{\dfrac{{{x}^{2}}}{x+1}}dx$
Đáp án B.
