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Câu hỏi: Tìm nguyên hàm $F\left( x \right)$ của hàm số $f\left( x \right)=2x+1-\dfrac{2}{x-2}$ biết $F\left( 1 \right)=3$.
A. $F\left( x \right)={{x}^{2}}+x-2\ln \left( 2-x \right)+1$.
B. $F\left( x \right)={{x}^{2}}+x+2\ln \left| x-2 \right|+1$.
C. $F\left( x \right)={{x}^{2}}+x-\ln \left| x-2 \right|+1$.
D. $F\left( x \right)={{x}^{2}}+x-2\ln \left| x-2 \right|+1$.
A. $F\left( x \right)={{x}^{2}}+x-2\ln \left( 2-x \right)+1$.
B. $F\left( x \right)={{x}^{2}}+x+2\ln \left| x-2 \right|+1$.
C. $F\left( x \right)={{x}^{2}}+x-\ln \left| x-2 \right|+1$.
D. $F\left( x \right)={{x}^{2}}+x-2\ln \left| x-2 \right|+1$.
$F\left( x \right)=\int{f\left( x \right)} \text{d}x=\int{\left( 2x+1-\dfrac{2}{x-2} \right)\text{d}x}={{x}^{2}}+x-2\ln \left| x-2 \right|+C$.
Mà $F\left( 1 \right)=3$ nên $C=1\Rightarrow $ $F\left( x \right)={{x}^{2}}+x-2\ln \left| x-2 \right|+1$.
Mà $F\left( 1 \right)=3$ nên $C=1\Rightarrow $ $F\left( x \right)={{x}^{2}}+x-2\ln \left| x-2 \right|+1$.
Đáp án D.