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Câu hỏi: Giả sử tích phân $I=\int\limits_{1}^{5}{\dfrac{1}{1+\sqrt{3\text{x}+1}}d\text{x}}=a+b.\ln 3+c.\ln 5$. Lúc đó:
A. $a+b+c=\dfrac{4}{3}$
B. $a+b+c=\dfrac{5}{3}$
C. $a+b+c=\dfrac{7}{3}$
D. $a+b+c=\dfrac{8}{3}$
A. $a+b+c=\dfrac{4}{3}$
B. $a+b+c=\dfrac{5}{3}$
C. $a+b+c=\dfrac{7}{3}$
D. $a+b+c=\dfrac{8}{3}$
Đặt $t=\sqrt{3x+1}\Rightarrow {{t}^{2}}=3\text{x}+1\Rightarrow 2t\text{d}t=3\text{dx}$. Đổi cận $\left| \begin{aligned}
& x=1\Rightarrow t=2 \\
& x=5\Rightarrow t=4 \\
\end{aligned} \right.$.
Khi đó $I=\dfrac{2}{3}\int\limits_{2}^{4}{\dfrac{t\text{d}t}{1+t}}=\dfrac{2}{3}\int\limits_{2}^{4}{\left( 1-\dfrac{1}{t+1} \right)dt}=\left. \dfrac{2}{3}\left( t-\ln \left| t+1 \right| \right) \right|_{2}^{4}=\dfrac{4}{3}-\dfrac{2}{3}\ln \dfrac{5}{3}=\dfrac{4}{3}-\dfrac{2}{3}\ln 5+\dfrac{2}{3}\ln 3$.
Do đó $a+b+c=\dfrac{4}{3}-\dfrac{2}{3}+\dfrac{2}{3}=\dfrac{4}{3}$.
& x=1\Rightarrow t=2 \\
& x=5\Rightarrow t=4 \\
\end{aligned} \right.$.
Khi đó $I=\dfrac{2}{3}\int\limits_{2}^{4}{\dfrac{t\text{d}t}{1+t}}=\dfrac{2}{3}\int\limits_{2}^{4}{\left( 1-\dfrac{1}{t+1} \right)dt}=\left. \dfrac{2}{3}\left( t-\ln \left| t+1 \right| \right) \right|_{2}^{4}=\dfrac{4}{3}-\dfrac{2}{3}\ln \dfrac{5}{3}=\dfrac{4}{3}-\dfrac{2}{3}\ln 5+\dfrac{2}{3}\ln 3$.
Do đó $a+b+c=\dfrac{4}{3}-\dfrac{2}{3}+\dfrac{2}{3}=\dfrac{4}{3}$.
Đáp án A.