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Câu hỏi: Một nguyên hàm của hàm số $f\left( x \right)=\dfrac{{{x}^{2}}+1}{{{x}^{4}}+2{{x}^{3}}-10{{x}^{2}}-2x+1}$ có dạng $F\left( x \right)=\dfrac{a}{b}\ln \left| \dfrac{{{x}^{2}}-cx-1}{{{x}^{2}}+dx-1} \right|,$ trong đó $a, b, c, d$ là các số nguyên dương và phân số $\dfrac{a}{b}$ tối giản. Tính $a+b+c+d.$
A. $24.$
B. $21.$
C. $15.$
D. $13.$
A. $24.$
B. $21.$
C. $15.$
D. $13.$
$\int{\dfrac{{{x}^{2}}+1}{{{x}^{4}}+2{{x}^{3}}-10{{x}^{2}}-2x+1}\text{d}x}=\int{\dfrac{1+\dfrac{1}{{{x}^{2}}}}{{{x}^{2}}+2x-10-2\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}}\text{d}x}=\int{\dfrac{\text{d}\left( x-\dfrac{1}{x} \right)}{{{\left( x-\dfrac{1}{x} \right)}^{2}}+2\left( x-\dfrac{1}{x} \right)-8}}$
$=\dfrac{1}{6}\int{\dfrac{\left( x-\dfrac{1}{x}+4 \right)-\left( x-\dfrac{1}{x}-2 \right)}{\left( x-\dfrac{1}{x}-2 \right)\left( x-\dfrac{1}{x}+4 \right)}\text{d}\left( x-\dfrac{1}{x} \right)}=\dfrac{1}{6}\int{\left[ \dfrac{\text{d}\left( x-\dfrac{1}{x} \right)}{x-\dfrac{1}{x}-2}-\dfrac{\text{d}\left( x-\dfrac{1}{x} \right)}{x-\dfrac{1}{x}+4} \right]}$
$=\dfrac{1}{6}\ln \left| x-\dfrac{1}{x}-2 \right|-\dfrac{1}{6}\ln \left| x-\dfrac{1}{x}+4 \right|+C=\dfrac{1}{6}\ln \left| \dfrac{x-\dfrac{1}{x}-2}{x-\dfrac{1}{x}+4} \right|+C=\dfrac{1}{6}\ln \left| \dfrac{{{x}^{2}}-2x-1}{{{x}^{2}}+4x-1} \right|+C$.
$=\dfrac{1}{6}\int{\dfrac{\left( x-\dfrac{1}{x}+4 \right)-\left( x-\dfrac{1}{x}-2 \right)}{\left( x-\dfrac{1}{x}-2 \right)\left( x-\dfrac{1}{x}+4 \right)}\text{d}\left( x-\dfrac{1}{x} \right)}=\dfrac{1}{6}\int{\left[ \dfrac{\text{d}\left( x-\dfrac{1}{x} \right)}{x-\dfrac{1}{x}-2}-\dfrac{\text{d}\left( x-\dfrac{1}{x} \right)}{x-\dfrac{1}{x}+4} \right]}$
$=\dfrac{1}{6}\ln \left| x-\dfrac{1}{x}-2 \right|-\dfrac{1}{6}\ln \left| x-\dfrac{1}{x}+4 \right|+C=\dfrac{1}{6}\ln \left| \dfrac{x-\dfrac{1}{x}-2}{x-\dfrac{1}{x}+4} \right|+C=\dfrac{1}{6}\ln \left| \dfrac{{{x}^{2}}-2x-1}{{{x}^{2}}+4x-1} \right|+C$.
Đáp án D.