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Câu hỏi: Cho $\int\limits_{0}^{1}{\dfrac{dx}{{{e}^{5x}}+2}}=a+b\ln 3+c\ln \left( {{e}^{5}}+2 \right)$ với $a,b,c$ là các số hữu tỉ. Giá trị của $4a+5b+5c$ bằng
A. $2$.
B. $0$.
C. $-2$.
D. $3$.
Ta có $\int\limits_{0}^{1}{\dfrac{dx}{{{e}^{5x}}+2}}=\dfrac{1}{5}\int\limits_{0}^{1}{\dfrac{d{{e}^{5x}}}{\left( {{e}^{5x}}+2 \right){{e}^{5x}}}}$ $=\dfrac{1}{10}\int\limits_{0}^{1}{\left( \dfrac{1}{{{e}^{5x}}}-\dfrac{1}{{{e}^{5x}}+2} \right)}d{{e}^{5x}}$ $=\left. \dfrac{1}{10}\left( \ln {{e}^{5x}}-\ln \left( {{e}^{5x}}+2 \right) \right) \right|_{0}^{1}$
$=\dfrac{1}{2}-\dfrac{1}{10}\ln \left( {{e}^{5}}+2 \right)+\dfrac{1}{10}\ln 3$. $\Rightarrow a=\dfrac{1}{2},$ $b=\dfrac{1}{10},$ $c=-\dfrac{1}{10}$.
Vậy $4a+5b+5c=2$
A. $2$.
B. $0$.
C. $-2$.
D. $3$.
Ta có $\int\limits_{0}^{1}{\dfrac{dx}{{{e}^{5x}}+2}}=\dfrac{1}{5}\int\limits_{0}^{1}{\dfrac{d{{e}^{5x}}}{\left( {{e}^{5x}}+2 \right){{e}^{5x}}}}$ $=\dfrac{1}{10}\int\limits_{0}^{1}{\left( \dfrac{1}{{{e}^{5x}}}-\dfrac{1}{{{e}^{5x}}+2} \right)}d{{e}^{5x}}$ $=\left. \dfrac{1}{10}\left( \ln {{e}^{5x}}-\ln \left( {{e}^{5x}}+2 \right) \right) \right|_{0}^{1}$
$=\dfrac{1}{2}-\dfrac{1}{10}\ln \left( {{e}^{5}}+2 \right)+\dfrac{1}{10}\ln 3$. $\Rightarrow a=\dfrac{1}{2},$ $b=\dfrac{1}{10},$ $c=-\dfrac{1}{10}$.
Vậy $4a+5b+5c=2$
Đáp án A.