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Câu hỏi: Họ các nguyên hàm $\int{\dfrac{1}{{{\left( 2x-1 \right)}^{2}}}\text{dx}}$ là
A. $\dfrac{-1}{4x-2}+C$.
B. $\dfrac{1}{2x-1}+C$.
C. $\dfrac{-1}{2x-1}+C$.
D. $\dfrac{1}{4x-2}+C$.
$\int{\dfrac{1}{{{\left( 2x-1 \right)}^{2}}}\text{dx}}=\dfrac{1}{2}\int{{{\left( 2x-1 \right)}^{-2}}\text{d}\left( 2x-1 \right)}=\dfrac{1}{2}.\dfrac{{{\left( 2x-1 \right)}^{-1}}}{-1}+C=\dfrac{-1}{4x-2}+C$.
A. $\dfrac{-1}{4x-2}+C$.
B. $\dfrac{1}{2x-1}+C$.
C. $\dfrac{-1}{2x-1}+C$.
D. $\dfrac{1}{4x-2}+C$.
$\int{\dfrac{1}{{{\left( 2x-1 \right)}^{2}}}\text{dx}}=\dfrac{1}{2}\int{{{\left( 2x-1 \right)}^{-2}}\text{d}\left( 2x-1 \right)}=\dfrac{1}{2}.\dfrac{{{\left( 2x-1 \right)}^{-1}}}{-1}+C=\dfrac{-1}{4x-2}+C$.
Đáp án A.