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Câu hỏi: Họ các nguyên hàm của hàm số $f(x)=\dfrac{1}{x{{(x+1)}^{2}}}$ trên khoảng $\left( 0;+\infty \right)$ là
A. $\ln \left( \dfrac{x}{x+1} \right)-\dfrac{1}{x+1}+C$.
B. $-\ln \left( \dfrac{x}{x+1} \right)+\dfrac{1}{x+1}+C.$
C. $\ln \left( \dfrac{x}{x+1} \right)+\dfrac{1}{x+1}+C$.
D. $-\ln \left( \dfrac{x}{x+1} \right)-\dfrac{1}{x+1}+C.$
A. $\ln \left( \dfrac{x}{x+1} \right)-\dfrac{1}{x+1}+C$.
B. $-\ln \left( \dfrac{x}{x+1} \right)+\dfrac{1}{x+1}+C.$
C. $\ln \left( \dfrac{x}{x+1} \right)+\dfrac{1}{x+1}+C$.
D. $-\ln \left( \dfrac{x}{x+1} \right)-\dfrac{1}{x+1}+C.$
$\int{f(x)} dx=\int{\dfrac{1}{x{{(x+1)}^{2}}}dx}=\int{\left( \dfrac{1}{x}-\dfrac{1}{x+1}-\dfrac{1}{{{(x+1)}^{2}}} \right)dx}$ $=\ln \dfrac{x}{x+1}+\dfrac{1}{x+1}+C, \forall x\in \left( 0;+\infty \right)$.
Đáp án C.