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Câu hỏi: Giá trị của tích phân $I=\int\limits_{e}^{{{e}^{2}}}{\left( \dfrac{1+x+{{x}^{2}}}{x} \right)}dx=a$. Biểu thức $P=a-1$ có giá trị là:
A. $P=e+\dfrac{1}{2}{{e}^{2}}+\dfrac{1}{2}{{e}^{4}}$
B. $P=-e+\dfrac{1}{2}{{e}^{2}}+\dfrac{1}{2}{{e}^{4}}$
C. $P=-e-\dfrac{1}{2}{{e}^{2}}+\dfrac{1}{2}{{e}^{4}}$
D. $P=e+\dfrac{1}{2}{{e}^{2}}-\dfrac{1}{2}{{e}^{4}}$
A. $P=e+\dfrac{1}{2}{{e}^{2}}+\dfrac{1}{2}{{e}^{4}}$
B. $P=-e+\dfrac{1}{2}{{e}^{2}}+\dfrac{1}{2}{{e}^{4}}$
C. $P=-e-\dfrac{1}{2}{{e}^{2}}+\dfrac{1}{2}{{e}^{4}}$
D. $P=e+\dfrac{1}{2}{{e}^{2}}-\dfrac{1}{2}{{e}^{4}}$
Giá trị của tích phân $I=\int\limits_{e}^{{{e}^{2}}}{\left( \dfrac{1+x+{{x}^{2}}}{x} \right)}dx=a$. Biểu thức $P=a-1$ có giá trị là:
Ta có:
$I=\int\limits_{e}^{{{e}^{2}}}{\left( \dfrac{1+x+{{x}^{2}}}{x} \right)}dx=\int\limits_{e}^{{{e}^{2}}}{\left( \dfrac{1}{x}+1+x \right)}dx=\left. \left( \ln \left| x \right|+x+\dfrac{{{x}^{2}}}{2} \right) \right|_{e}^{{{e}^{2}}}=1-e+\dfrac{{{e}^{2}}}{2}+\dfrac{{{e}^{4}}}{2}$.
$\Rightarrow a=1-e+\dfrac{{{e}^{2}}}{2}+\dfrac{{{e}^{4}}}{2}\Leftrightarrow a-1=-e+\dfrac{{{e}^{2}}}{2}+\dfrac{{{e}^{4}}}{2}\Leftrightarrow P=-e+\dfrac{{{e}^{2}}}{2}+\dfrac{{{e}^{4}}}{2}$.
Ta có:
$I=\int\limits_{e}^{{{e}^{2}}}{\left( \dfrac{1+x+{{x}^{2}}}{x} \right)}dx=\int\limits_{e}^{{{e}^{2}}}{\left( \dfrac{1}{x}+1+x \right)}dx=\left. \left( \ln \left| x \right|+x+\dfrac{{{x}^{2}}}{2} \right) \right|_{e}^{{{e}^{2}}}=1-e+\dfrac{{{e}^{2}}}{2}+\dfrac{{{e}^{4}}}{2}$.
$\Rightarrow a=1-e+\dfrac{{{e}^{2}}}{2}+\dfrac{{{e}^{4}}}{2}\Leftrightarrow a-1=-e+\dfrac{{{e}^{2}}}{2}+\dfrac{{{e}^{4}}}{2}\Leftrightarrow P=-e+\dfrac{{{e}^{2}}}{2}+\dfrac{{{e}^{4}}}{2}$.
Đáp án B.