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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục và có đạo hàm trên đoạn $\left[ -\dfrac{1}{2};\dfrac{1}{2} \right]$ thỏa mãn $\int\limits_{\dfrac{-1}{2}}^{\dfrac{1}{2}}{\left[ {{f}^{2}}\left( x \right)-2f\left( x \right).\left( 3-x \right) \right]\text{d}x}=\dfrac{-109}{12}.$ Tính tích phân $I=\int\limits_{0}^{\dfrac{1}{2}}{\dfrac{f\left( x \right)}{{{x}^{2}}-1}\text{d}x}.$
A. $I=\ln \dfrac{7}{9}.$
B. $I=\ln \dfrac{2}{9}.$
C. $I=\ln \dfrac{5}{9}.$
D. $I=\ln \dfrac{8}{9}.$
A. $I=\ln \dfrac{7}{9}.$
B. $I=\ln \dfrac{2}{9}.$
C. $I=\ln \dfrac{5}{9}.$
D. $I=\ln \dfrac{8}{9}.$
Ta tính được $\int\limits_{-\dfrac{1}{2}}^{\dfrac{1}{2}}{{{\left( 3-x \right)}^{2}}d\text{x}}=\dfrac{-109}{12}$.
Do đó $\int\limits_{-\dfrac{1}{1}}^{\dfrac{1}{2}}{\left[ {{f}^{2}}\left( x \right)-2f\left( x \right).\left( 3-x \right) \right]d\text{x}}=-\int\limits_{-\dfrac{1}{2}}^{\dfrac{1}{2}}{{{\left( 3-x \right)}^{2}}d\text{x}}$
$\Leftrightarrow \int\limits_{-\dfrac{1}{2}}^{\dfrac{1}{2}}{{{\left[ f\left( x \right)-\left( 3-x \right) \right]}^{2}}d\text{x}}=0$
$\Leftrightarrow f\left( x \right)=3-x\Rightarrow I=\int\limits_{0}^{\dfrac{1}{2}}{\dfrac{3-x}{{{x}^{2}}-1}d\text{x}}=\ln \dfrac{2}{9}$.
Do đó $\int\limits_{-\dfrac{1}{1}}^{\dfrac{1}{2}}{\left[ {{f}^{2}}\left( x \right)-2f\left( x \right).\left( 3-x \right) \right]d\text{x}}=-\int\limits_{-\dfrac{1}{2}}^{\dfrac{1}{2}}{{{\left( 3-x \right)}^{2}}d\text{x}}$
$\Leftrightarrow \int\limits_{-\dfrac{1}{2}}^{\dfrac{1}{2}}{{{\left[ f\left( x \right)-\left( 3-x \right) \right]}^{2}}d\text{x}}=0$
$\Leftrightarrow f\left( x \right)=3-x\Rightarrow I=\int\limits_{0}^{\dfrac{1}{2}}{\dfrac{3-x}{{{x}^{2}}-1}d\text{x}}=\ln \dfrac{2}{9}$.
Đáp án B.