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Câu hỏi: Biết $\int\limits_{1}^{5}{\dfrac{1}{1+\sqrt{3x+1}}}\text{d}x=a+b\ln 3+c\ln 5$ $\left( a,b,c\in Q \right)$. Giá trị của $a+2b+3c$ bằng:
A. $\dfrac{2}{3}$.
B. $\dfrac{5}{3}$.
C. $\dfrac{8}{3}$.
D. $\dfrac{7}{3}$.
A. $\dfrac{2}{3}$.
B. $\dfrac{5}{3}$.
C. $\dfrac{8}{3}$.
D. $\dfrac{7}{3}$.
Đặt $t=\sqrt{3x+1}\Rightarrow {{t}^{2}}=3x+1\Rightarrow \dfrac{2}{3}t \text{d}t= \text{d}x$.
Đổi cận $\left\{ \begin{aligned}
& x=1\Rightarrow t=2 \\
& x=5\Rightarrow t=4. \\
\end{aligned} \right.$
Do đó $\int\limits_{1}^{5}{\dfrac{1}{1+\sqrt{3x+1}}} \text{d}x=\dfrac{2}{3}\int\limits_{2}^{4}{\dfrac{t}{1+t}} \text{d}t=\dfrac{2}{3}\int\limits_{2}^{4}{\left( 1-\dfrac{1}{1+t} \right)} \text{d}t=\dfrac{2}{3}\left. \left( t-\ln \left( 1+t \right) \right) \right|_{2}^{4}=\dfrac{4}{3}+\dfrac{2}{3}\ln 3-\dfrac{2}{3}\ln 5$.
Suy ra $a=\dfrac{4}{3}; b=\dfrac{2}{3}; c=-\dfrac{2}{3}$.
Vậy $a+2b+3c=\dfrac{2}{3}$.
Đổi cận $\left\{ \begin{aligned}
& x=1\Rightarrow t=2 \\
& x=5\Rightarrow t=4. \\
\end{aligned} \right.$
Do đó $\int\limits_{1}^{5}{\dfrac{1}{1+\sqrt{3x+1}}} \text{d}x=\dfrac{2}{3}\int\limits_{2}^{4}{\dfrac{t}{1+t}} \text{d}t=\dfrac{2}{3}\int\limits_{2}^{4}{\left( 1-\dfrac{1}{1+t} \right)} \text{d}t=\dfrac{2}{3}\left. \left( t-\ln \left( 1+t \right) \right) \right|_{2}^{4}=\dfrac{4}{3}+\dfrac{2}{3}\ln 3-\dfrac{2}{3}\ln 5$.
Suy ra $a=\dfrac{4}{3}; b=\dfrac{2}{3}; c=-\dfrac{2}{3}$.
Vậy $a+2b+3c=\dfrac{2}{3}$.
Đáp án A.