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Câu hỏi: Nguyên hàm $\int{\dfrac{3x+3}{-{{x}^{2}}-x+2}dx}$ là:
A. $2\ln \left| x-1 \right|-\ln \left| x+2 \right|+C$.
B. $-2\ln \left| x-1 \right|+\ln \left| x+2 \right|+C$.
C. $2\ln \left| x-1 \right|+\ln \left| x+2 \right|+C$.
D. $-2\ln \left| x-1 \right|-\ln \left| x+2 \right|+C$.
A. $2\ln \left| x-1 \right|-\ln \left| x+2 \right|+C$.
B. $-2\ln \left| x-1 \right|+\ln \left| x+2 \right|+C$.
C. $2\ln \left| x-1 \right|+\ln \left| x+2 \right|+C$.
D. $-2\ln \left| x-1 \right|-\ln \left| x+2 \right|+C$.
Ta có:
$\int{\dfrac{3x+3}{-{{x}^{2}}-x+2}dx}=\int{\dfrac{3x+3}{\left( 1-x \right)\left( x+2 \right)}dx}=\int{\left( \dfrac{2}{1-x}-\dfrac{1}{x+2} \right)dx=-2\ln \left| x-1 \right|-\ln \left| x+2 \right|+C}$.
$\int{\dfrac{3x+3}{-{{x}^{2}}-x+2}dx}=\int{\dfrac{3x+3}{\left( 1-x \right)\left( x+2 \right)}dx}=\int{\left( \dfrac{2}{1-x}-\dfrac{1}{x+2} \right)dx=-2\ln \left| x-1 \right|-\ln \left| x+2 \right|+C}$.
Đáp án B.