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Câu hỏi: Họ tất cả các nguyên hàm của hàm số $f\left( x \right)=\dfrac{3x+1}{{{\left( x-1 \right)}^{2}}}$ trên khoảng $\left( 1; +\infty \right)$ là
A. $3\ln \left( x-1 \right)-\dfrac{2}{x-1}+C$
B. $3\ln \left( x-1 \right)+\dfrac{1}{x-1}+C$
C. $3\ln \left( x-1 \right)-\dfrac{1}{x-1}+C$
D. $3\ln \left( x-1 \right)+\dfrac{2}{x-1}+C$
A. $3\ln \left( x-1 \right)-\dfrac{2}{x-1}+C$
B. $3\ln \left( x-1 \right)+\dfrac{1}{x-1}+C$
C. $3\ln \left( x-1 \right)-\dfrac{1}{x-1}+C$
D. $3\ln \left( x-1 \right)+\dfrac{2}{x-1}+C$
Ta có $f\left( x \right)=\dfrac{3x-3+2}{{{\left( x-1 \right)}^{2}}}=\dfrac{3\left( x-1 \right)+2}{{{\left( x-1 \right)}^{2}}}=\dfrac{3}{x+1}+\dfrac{2}{{{\left( x-1 \right)}^{2}}}$
Vậy $\int{f\left( x \right)dx}=\int{\left( \dfrac{3}{x-1}+\dfrac{2}{{{\left( x-1 \right)}^{2}}} \right)dx}=3\ln \left( x-1 \right)-\dfrac{2}{x-1}+C$ vì $x>1$
Vậy $\int{f\left( x \right)dx}=\int{\left( \dfrac{3}{x-1}+\dfrac{2}{{{\left( x-1 \right)}^{2}}} \right)dx}=3\ln \left( x-1 \right)-\dfrac{2}{x-1}+C$ vì $x>1$
Đáp án A.