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Câu hỏi: Biết $\int\limits_{2}^{3}{\dfrac{-2{{x}^{2}}+14x}{{{x}^{2}}-1}dx}=a\ln 2+b\ln 3+c,\left( a,b,c\in \mathbb{Z} \right).$ Giá trị của ${{a}^{2}}+b+c$ bằng
A. 494
B. 484
C. 474
D. 464
A. 494
B. 484
C. 474
D. 464
Cách giải:
$\int\limits_{2}^{3}{\dfrac{-2{{x}^{2}}+14x}{{{x}^{2}}-1}dx}=\int\limits_{2}^{3}{\left( -2+\dfrac{14x-2}{{{x}^{2}}-1} \right)dx}=\int\limits_{2}^{3}{\left( -2+\dfrac{6}{x-1}+\dfrac{8}{x+1} \right)dx}=-2x+6\ln \left| x-1 \right|+8\ln \left| x+1 \right|\left| \begin{aligned}
& 3 \\
& 2 \\
\end{aligned} \right.=22\ln 2-8\ln 3-2$
$\Rightarrow a=22,b=-8,c=-2$
Do đó: ${{a}^{2}}+b+c={{22}^{2}}-8-2=474.$
$\int\limits_{2}^{3}{\dfrac{-2{{x}^{2}}+14x}{{{x}^{2}}-1}dx}=\int\limits_{2}^{3}{\left( -2+\dfrac{14x-2}{{{x}^{2}}-1} \right)dx}=\int\limits_{2}^{3}{\left( -2+\dfrac{6}{x-1}+\dfrac{8}{x+1} \right)dx}=-2x+6\ln \left| x-1 \right|+8\ln \left| x+1 \right|\left| \begin{aligned}
& 3 \\
& 2 \\
\end{aligned} \right.=22\ln 2-8\ln 3-2$
$\Rightarrow a=22,b=-8,c=-2$
Do đó: ${{a}^{2}}+b+c={{22}^{2}}-8-2=474.$
Đáp án C.