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Câu hỏi: Biết $\int\limits_{2}^{5}{\left( 2x+1 \right)\ln \left( {{x}^{2}}-1 \right)\text{d}x}=a\ln 3+b\ln 2-c$ với $a,b,c$ là các số nguyên. Khi đó ${{a}^{2}}+2b-{{c}^{2}}$ bằng
A. $8.$
B. $19.$
C. 6.
D. $5$.
A. $8.$
B. $19.$
C. 6.
D. $5$.
Đặt $\left\{ \begin{aligned}
& u=\ln \left( {{x}^{2}}-1 \right)\Rightarrow \text{d}u=\dfrac{2x}{{{x}^{2}}-1}\text{d}x \\
& \text{d}v=\left( 2x+1 \right)\text{d}x\Rightarrow v={{x}^{2}}+x \\
\end{aligned} \right.$
Khi đó $\int\limits_{2}^{5}{\left( 2x+1 \right)\ln \left( {{x}^{2}}-1 \right)\text{d}x}=\left. \left( {{x}^{2}}+x \right)\ln \left( {{x}^{2}}-1 \right) \right|_{2}^{5}-\int\limits_{2}^{5}{\left( {{x}^{2}}+x \right)\dfrac{2x}{{{x}^{2}}-1}\text{d}x}$
$=30\ln 24-6\ln 3-2\int\limits_{2}^{5}{\dfrac{{{x}^{2}}}{x-1}}\text{d}x=90\ln 2+24\ln 3-2\int\limits_{2}^{5}{\left( x+1+\dfrac{1}{x-1} \right)\text{d}x}$
$=90\ln 2+24\ln 3-27-4\ln 2=24\ln 3+86\ln 2-27\Rightarrow a=24,b=86,c=27\Rightarrow {{a}^{2}}+2b-{{c}^{2}}=19.$
& u=\ln \left( {{x}^{2}}-1 \right)\Rightarrow \text{d}u=\dfrac{2x}{{{x}^{2}}-1}\text{d}x \\
& \text{d}v=\left( 2x+1 \right)\text{d}x\Rightarrow v={{x}^{2}}+x \\
\end{aligned} \right.$
Khi đó $\int\limits_{2}^{5}{\left( 2x+1 \right)\ln \left( {{x}^{2}}-1 \right)\text{d}x}=\left. \left( {{x}^{2}}+x \right)\ln \left( {{x}^{2}}-1 \right) \right|_{2}^{5}-\int\limits_{2}^{5}{\left( {{x}^{2}}+x \right)\dfrac{2x}{{{x}^{2}}-1}\text{d}x}$
$=30\ln 24-6\ln 3-2\int\limits_{2}^{5}{\dfrac{{{x}^{2}}}{x-1}}\text{d}x=90\ln 2+24\ln 3-2\int\limits_{2}^{5}{\left( x+1+\dfrac{1}{x-1} \right)\text{d}x}$
$=90\ln 2+24\ln 3-27-4\ln 2=24\ln 3+86\ln 2-27\Rightarrow a=24,b=86,c=27\Rightarrow {{a}^{2}}+2b-{{c}^{2}}=19.$
Đáp án B.
