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Câu hỏi: Cho tích phân $\int\limits_{1}^{2}{\dfrac{x\ln x dx}{{{\left( {{x}^{2}}+1 \right)}^{2}}}}=a\ln 2+b\ln 3+c\ln 5$ (với a, b, c là các số hữu tỉ). Khi đó $a+b-c$ bằng
A. $\dfrac{2}{5}$
B. $-\dfrac{2}{5}$
C. $\dfrac{9}{10}$
D. $-\dfrac{9}{10}$
A. $\dfrac{2}{5}$
B. $-\dfrac{2}{5}$
C. $\dfrac{9}{10}$
D. $-\dfrac{9}{10}$
Đặt $\left\{ \begin{aligned}
& u=\ln x \\
& dv=\dfrac{x}{{{\left( {{x}^{2}}+1 \right)}^{2}}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{dx}{x} \\
& v=-\dfrac{1}{2\left( {{x}^{2}}+1 \right)} \\
\end{aligned} \right.$
Ta có $\int\limits_{1}^{2}{\dfrac{x\ln xdx}{{{\left( {{x}^{2}}+1 \right)}^{2}}}=\left. \left( -\dfrac{\operatorname{lnx}}{2\left( {{x}^{2}}+1 \right)} \right) \right|}_{1}^{2}+\int\limits_{1}^{2}{\underbrace{\dfrac{dx}{2x({{x}^{2}}+1)}}_{I}=-\dfrac{\ln 2}{10}+I}$
Tính $I=\int\limits_{1}^{2}{\dfrac{dx}{2x\left( {{x}^{2}}+1 \right)}}$, đặt $t={{x}^{2}}+1\Rightarrow dt=2xdx$, với $x=1\Rightarrow t=2;x=2\Rightarrow t=5$
Suy ra
$\begin{aligned}
& I=\int\limits_{1}^{2}{\dfrac{dx}{2x\left( {{x}^{2}}+1 \right)}=\int\limits_{1}^{2}{\dfrac{xdx}{2{{x}^{2}}\left( {{x}^{2}}+1 \right)}=\dfrac{1}{4}\int\limits_{2}^{5}{\dfrac{dt}{t\left( t-1 \right)}=\dfrac{1}{4}\int\limits_{2}^{5}{\dfrac{dt}{t-1}-\dfrac{1}{4}\int\limits_{2}^{5}{\dfrac{dt}{t}=\left. \dfrac{1}{4}\ln \left| t-1 \right| \right|}}}}}_{2}^{5}-\left. \dfrac{1}{4}\ln \left| t \right| \right|_{2}^{5} \\
& =\left. \dfrac{1}{4}\ln \left| t-1 \right| \right|_{2}^{5}-\left. \dfrac{1}{4}\ln \left| t \right| \right|_{2}^{5}=\dfrac{1}{4}\ln 4-\dfrac{1}{4}\ln 5+\dfrac{1}{4}\ln 2=\dfrac{3}{4}\ln 2-\dfrac{1}{4}\ln 5 \\
\end{aligned}$
Vậy $\int\limits_{1}^{2}{\dfrac{x\ln xdx}{{{\left( {{x}^{2}}+1 \right)}^{2}}}=-\dfrac{\ln 2}{10}+\dfrac{3}{4}\ln 2-\dfrac{1}{4}\ln 5=\dfrac{13}{20}\ln 2-\dfrac{1}{4}\ln 5}$
Từ đó ta có $a=\dfrac{13}{20};b=0;c=-\dfrac{1}{4}$
Suy ra $a+b-c=\dfrac{13}{20}+0+\dfrac{1}{4}=\dfrac{9}{10}$
& u=\ln x \\
& dv=\dfrac{x}{{{\left( {{x}^{2}}+1 \right)}^{2}}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{dx}{x} \\
& v=-\dfrac{1}{2\left( {{x}^{2}}+1 \right)} \\
\end{aligned} \right.$
Ta có $\int\limits_{1}^{2}{\dfrac{x\ln xdx}{{{\left( {{x}^{2}}+1 \right)}^{2}}}=\left. \left( -\dfrac{\operatorname{lnx}}{2\left( {{x}^{2}}+1 \right)} \right) \right|}_{1}^{2}+\int\limits_{1}^{2}{\underbrace{\dfrac{dx}{2x({{x}^{2}}+1)}}_{I}=-\dfrac{\ln 2}{10}+I}$
Tính $I=\int\limits_{1}^{2}{\dfrac{dx}{2x\left( {{x}^{2}}+1 \right)}}$, đặt $t={{x}^{2}}+1\Rightarrow dt=2xdx$, với $x=1\Rightarrow t=2;x=2\Rightarrow t=5$
Suy ra
$\begin{aligned}
& I=\int\limits_{1}^{2}{\dfrac{dx}{2x\left( {{x}^{2}}+1 \right)}=\int\limits_{1}^{2}{\dfrac{xdx}{2{{x}^{2}}\left( {{x}^{2}}+1 \right)}=\dfrac{1}{4}\int\limits_{2}^{5}{\dfrac{dt}{t\left( t-1 \right)}=\dfrac{1}{4}\int\limits_{2}^{5}{\dfrac{dt}{t-1}-\dfrac{1}{4}\int\limits_{2}^{5}{\dfrac{dt}{t}=\left. \dfrac{1}{4}\ln \left| t-1 \right| \right|}}}}}_{2}^{5}-\left. \dfrac{1}{4}\ln \left| t \right| \right|_{2}^{5} \\
& =\left. \dfrac{1}{4}\ln \left| t-1 \right| \right|_{2}^{5}-\left. \dfrac{1}{4}\ln \left| t \right| \right|_{2}^{5}=\dfrac{1}{4}\ln 4-\dfrac{1}{4}\ln 5+\dfrac{1}{4}\ln 2=\dfrac{3}{4}\ln 2-\dfrac{1}{4}\ln 5 \\
\end{aligned}$
Vậy $\int\limits_{1}^{2}{\dfrac{x\ln xdx}{{{\left( {{x}^{2}}+1 \right)}^{2}}}=-\dfrac{\ln 2}{10}+\dfrac{3}{4}\ln 2-\dfrac{1}{4}\ln 5=\dfrac{13}{20}\ln 2-\dfrac{1}{4}\ln 5}$
Từ đó ta có $a=\dfrac{13}{20};b=0;c=-\dfrac{1}{4}$
Suy ra $a+b-c=\dfrac{13}{20}+0+\dfrac{1}{4}=\dfrac{9}{10}$
Đáp án C.