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Câu hỏi: Tìm họ nguyên hàm của hàm số $y=\dfrac{1}{{{\left( x+1 \right)}^{2}}}$.
A. $\int{\dfrac{d\text{x}}{{{\left( x+1 \right)}^{2}}}}=\dfrac{2}{{{\left( x+1 \right)}^{3}}}+C$
B. $\int{\dfrac{d\text{x}}{{{\left( x+1 \right)}^{2}}}}=-\dfrac{1}{x+1}+C$
C. $\int{\dfrac{d\text{x}}{{{\left( x+1 \right)}^{2}}}}=\dfrac{1}{x+1}+C$
D. $\int{\dfrac{d\text{x}}{{{\left( x+1 \right)}^{2}}}}=-\dfrac{2}{{{\left( x+1 \right)}^{3}}}+C$
A. $\int{\dfrac{d\text{x}}{{{\left( x+1 \right)}^{2}}}}=\dfrac{2}{{{\left( x+1 \right)}^{3}}}+C$
B. $\int{\dfrac{d\text{x}}{{{\left( x+1 \right)}^{2}}}}=-\dfrac{1}{x+1}+C$
C. $\int{\dfrac{d\text{x}}{{{\left( x+1 \right)}^{2}}}}=\dfrac{1}{x+1}+C$
D. $\int{\dfrac{d\text{x}}{{{\left( x+1 \right)}^{2}}}}=-\dfrac{2}{{{\left( x+1 \right)}^{3}}}+C$
Ta có $\int{\dfrac{d\text{x}}{{{\left( x+1 \right)}^{2}}}}=-\dfrac{1}{x+1}+C$.
Đáp án B.