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Câu hỏi: Cho $I=\int\limits_{0}^{1}{x\ln \left( 2+{{x}^{2}} \right)dx=a\ln 3+b\ln 2+c}$ với a, b, c là các số hữu tỷ. Giá trị của $a+b+c$ bằng
A. – 1.
B. 1.
C. 0.
D. 2.
A. – 1.
B. 1.
C. 0.
D. 2.
Đặt$\left\{ \begin{aligned}
& u=\ln \left( 2+{{x}^{2}} \right) \\
& dv=xdx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{2x}{2+{{x}^{2}}}dx \\
& v=\dfrac{{{x}^{2}}}{2} \\
\end{aligned} \right..$
Khi đó, $I=\int\limits_{0}^{1}{x\ln \left( 2+{{x}^{2}} \right)dx=\left. \left[ \dfrac{{{x}^{2}}}{2}\ln \left( 2+{{x}^{2}} \right) \right] \right|}_{0}^{1}-\int\limits_{0}^{1}{\dfrac{{{x}^{2}}}{2}.\dfrac{2x}{2+{{x}^{2}}}dx}$
$=\left. \left[ \dfrac{{{x}^{2}}}{2}\ln \left( 2+{{x}^{2}} \right) \right] \right|_{0}^{1}-\int\limits_{0}^{1}{\dfrac{{{x}^{3}}}{2+{{x}^{2}}}dx}$ (1).
Xét ${{I}_{1}}=\int\limits_{0}^{1}{\dfrac{{{x}^{3}}}{2+{{x}^{2}}}dx}=\int\limits_{0}^{1}{\left( x-\dfrac{2x}{{{x}^{2}}+2} \right)dx=\int\limits_{0}^{1}{xdx-\int\limits_{0}^{1}{\dfrac{2x}{{{x}^{2}}+2}xdx}}}$
$=\left. \dfrac{{{x}^{2}}}{2} \right|_{0}^{1}-\int\limits_{0}^{1}{\dfrac{d\left( {{x}^{2}}+2 \right)}{{{x}^{2}}+2}}=\left. \dfrac{{{x}^{2}}}{2} \right|_{0}^{1}-\left. \ln \left( {{x}^{2}}+2 \right) \right|_{0}^{1}=\dfrac{1}{2}-\ln 3+\ln 2.$
Thay vào (1), suy ra $I=\dfrac{1}{2}\ln 3-\dfrac{1}{2}+\ln 3-\ln 2=\dfrac{3}{2}\ln 3-\ln 2-\dfrac{1}{2}.$
Vậy$\left\{ \begin{aligned}
& a=\dfrac{3}{2} \\
& b=-1 \\
& c=-\dfrac{1}{2} \\
\end{aligned} \right.\Rightarrow a+b+c=0.$
& u=\ln \left( 2+{{x}^{2}} \right) \\
& dv=xdx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{2x}{2+{{x}^{2}}}dx \\
& v=\dfrac{{{x}^{2}}}{2} \\
\end{aligned} \right..$
Khi đó, $I=\int\limits_{0}^{1}{x\ln \left( 2+{{x}^{2}} \right)dx=\left. \left[ \dfrac{{{x}^{2}}}{2}\ln \left( 2+{{x}^{2}} \right) \right] \right|}_{0}^{1}-\int\limits_{0}^{1}{\dfrac{{{x}^{2}}}{2}.\dfrac{2x}{2+{{x}^{2}}}dx}$
$=\left. \left[ \dfrac{{{x}^{2}}}{2}\ln \left( 2+{{x}^{2}} \right) \right] \right|_{0}^{1}-\int\limits_{0}^{1}{\dfrac{{{x}^{3}}}{2+{{x}^{2}}}dx}$ (1).
Xét ${{I}_{1}}=\int\limits_{0}^{1}{\dfrac{{{x}^{3}}}{2+{{x}^{2}}}dx}=\int\limits_{0}^{1}{\left( x-\dfrac{2x}{{{x}^{2}}+2} \right)dx=\int\limits_{0}^{1}{xdx-\int\limits_{0}^{1}{\dfrac{2x}{{{x}^{2}}+2}xdx}}}$
$=\left. \dfrac{{{x}^{2}}}{2} \right|_{0}^{1}-\int\limits_{0}^{1}{\dfrac{d\left( {{x}^{2}}+2 \right)}{{{x}^{2}}+2}}=\left. \dfrac{{{x}^{2}}}{2} \right|_{0}^{1}-\left. \ln \left( {{x}^{2}}+2 \right) \right|_{0}^{1}=\dfrac{1}{2}-\ln 3+\ln 2.$
Thay vào (1), suy ra $I=\dfrac{1}{2}\ln 3-\dfrac{1}{2}+\ln 3-\ln 2=\dfrac{3}{2}\ln 3-\ln 2-\dfrac{1}{2}.$
Vậy$\left\{ \begin{aligned}
& a=\dfrac{3}{2} \\
& b=-1 \\
& c=-\dfrac{1}{2} \\
\end{aligned} \right.\Rightarrow a+b+c=0.$
Đáp án C.