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Câu hỏi: Mệnh đề nào sau đây đúng?
A. $\int\limits_{{}}^{{}}{\dfrac{1}{1-4x}dx}=-\dfrac{1}{4}.\ln \left| 8x-2 \right|+C.$
B. $\int\limits_{{}}^{{}}{\dfrac{1}{1-4x}dx}=\ln \left| 1-4x \right|+C.$
C. $\int\limits_{{}}^{{}}{\dfrac{1}{1-4x}dx}=-\dfrac{1}{4}.\ln \left| 1-4x \right|+C.$
D. $\int\limits_{{}}^{{}}{\dfrac{1}{1-4x}dx}=-4.\ln \dfrac{1}{\left| 1-4x \right|}+C.$
A. $\int\limits_{{}}^{{}}{\dfrac{1}{1-4x}dx}=-\dfrac{1}{4}.\ln \left| 8x-2 \right|+C.$
B. $\int\limits_{{}}^{{}}{\dfrac{1}{1-4x}dx}=\ln \left| 1-4x \right|+C.$
C. $\int\limits_{{}}^{{}}{\dfrac{1}{1-4x}dx}=-\dfrac{1}{4}.\ln \left| 1-4x \right|+C.$
D. $\int\limits_{{}}^{{}}{\dfrac{1}{1-4x}dx}=-4.\ln \dfrac{1}{\left| 1-4x \right|}+C.$
Đặt $u=1-4x\Rightarrow du=-4.dx\Rightarrow dx=-\dfrac{1}{4}du.$
$\int\limits_{{}}^{{}}{\dfrac{1}{1-4x}dx}=\int\limits_{{}}^{{}}{\dfrac{1}{u}\left( -\dfrac{1}{4}du \right)}=-\dfrac{1}{4}\int\limits_{{}}^{{}}{\dfrac{1}{u}du}=-\dfrac{1}{4}\ln \left| u \right|+C=-\dfrac{1}{4}.\ln \left| 1-4x \right|+C.$
$\int\limits_{{}}^{{}}{\dfrac{1}{1-4x}dx}=\int\limits_{{}}^{{}}{\dfrac{1}{u}\left( -\dfrac{1}{4}du \right)}=-\dfrac{1}{4}\int\limits_{{}}^{{}}{\dfrac{1}{u}du}=-\dfrac{1}{4}\ln \left| u \right|+C=-\dfrac{1}{4}.\ln \left| 1-4x \right|+C.$
Đáp án C.