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