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Câu hỏi: Gọi $a, b$ là các số nguyên dương nhỏ nhất sao cho $\int\limits_{0}^{1}{\dfrac{\text{d}x}{4-{{x}^{2}}}}=\dfrac{\ln a}{b}$. Giá trị của $a+b$ bằng
A. $5$.
B. $7$.
C. $6$.
D. $12$.
Ta có: $\int\limits_{0}^{1}{\dfrac{\text{d}x}{4-{{x}^{2}}}}=\int\limits_{0}^{1}{\dfrac{\text{d}x}{(2-x)(2+x)}}=\dfrac{1}{4}\int\limits_{0}^{1}{\left( \dfrac{1}{2-x}+\dfrac{1}{2+x} \right)\text{d}x}=\left. \dfrac{1}{4}\ln \left( \dfrac{x+2}{2-x} \right) \right|_{0}^{1}=\dfrac{1}{4}\ln 3$
$\Rightarrow a=3,b=4$ $\Rightarrow a+b=7$
A. $5$.
B. $7$.
C. $6$.
D. $12$.
Ta có: $\int\limits_{0}^{1}{\dfrac{\text{d}x}{4-{{x}^{2}}}}=\int\limits_{0}^{1}{\dfrac{\text{d}x}{(2-x)(2+x)}}=\dfrac{1}{4}\int\limits_{0}^{1}{\left( \dfrac{1}{2-x}+\dfrac{1}{2+x} \right)\text{d}x}=\left. \dfrac{1}{4}\ln \left( \dfrac{x+2}{2-x} \right) \right|_{0}^{1}=\dfrac{1}{4}\ln 3$
$\Rightarrow a=3,b=4$ $\Rightarrow a+b=7$
Đáp án B.