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Câu hỏi: Tích phân $\int\limits_{0}^{1}{\dfrac{1}{{{x}^{2}}+4x+3}\text{d}x}$ có kết quả là
A. $\dfrac{1}{3}\ln \dfrac{3}{2}$.
B. $\dfrac{1}{2}\ln \dfrac{3}{2}$.
C. $-\dfrac{1}{2}\ln \dfrac{3}{2}$.
D. $\ln \dfrac{3}{2}$.
A. $\dfrac{1}{3}\ln \dfrac{3}{2}$.
B. $\dfrac{1}{2}\ln \dfrac{3}{2}$.
C. $-\dfrac{1}{2}\ln \dfrac{3}{2}$.
D. $\ln \dfrac{3}{2}$.
Ta có $\int\limits_{0}^{1}{\dfrac{1}{{{x}^{2}}+4x+3}\text{d}x}=\int\limits_{0}^{1}{\dfrac{1}{\left( x+1 \right)\left( x+3 \right)}\text{d}x}=\dfrac{1}{2}\int\limits_{0}^{1}{\left( \dfrac{1}{x+1}-\dfrac{1}{x+3} \right)\text{d}x}=\dfrac{1}{2}\ln \dfrac{x+1}{x+3}\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.$
$=\dfrac{1}{2}\left( \ln \dfrac{1}{2}-\ln \dfrac{1}{3} \right)=\dfrac{1}{2}\ln \dfrac{3}{2}$
& 1 \\
& 0 \\
\end{aligned} \right.$
$=\dfrac{1}{2}\left( \ln \dfrac{1}{2}-\ln \dfrac{1}{3} \right)=\dfrac{1}{2}\ln \dfrac{3}{2}$
Đáp án B.