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Câu hỏi: Biết rằng $\int\limits_{0}^{1}{\dfrac{dx}{3x+5\sqrt{3x+1}+7}=a\ln 2+b\ln 3+c\ln 5,}$ với $a,b,c$ là các số hữu tỉ. Giá trị của $a+b+c$ bằng
A. $-\dfrac{10}{3}$
B. $-\dfrac{5}{3}$
C. $\dfrac{10}{3}$
D. $\dfrac{5}{3}$
A. $-\dfrac{10}{3}$
B. $-\dfrac{5}{3}$
C. $\dfrac{10}{3}$
D. $\dfrac{5}{3}$
HD: Đặt $t=\sqrt{3x+1}\Rightarrow {{t}^{2}}=3x+1\Rightarrow 2tdt=3dx\Rightarrow dx=\dfrac{2t}{3}dt$
Đổi cận $\left| \begin{aligned}
& x=0\Rightarrow t=1 \\
& x=1\Rightarrow t=2 \\
\end{aligned} \right.\Rightarrow I=\dfrac{2}{3}\int\limits_{1}^{2}{\dfrac{tdt}{{{t}^{2}}+5t+6}=\dfrac{2}{3}\int\limits_{1}^{2}{\dfrac{t}{\left( t+2 \right)\left( t+3 \right)}dt=\dfrac{2}{3}\int\limits_{1}^{2}{\dfrac{3\left( t+2 \right)-2\left( t+3 \right)}{\left( t+2 \right)\left( t+3 \right)}dt}}}$
$=\dfrac{2}{3}\int\limits_{1}^{2}{\left( \dfrac{3}{t+3}-\dfrac{2}{t+2} \right)dt=2\ln \left| t+3 \right|}\left| _{1}^{2} \right.-\dfrac{4}{3}\ln \left| t+2 \right|\left| _{1}^{2} \right.=2\ln \dfrac{5}{4}-\dfrac{4}{3}\ln \dfrac{4}{3}=2\ln 5-\dfrac{20}{3}\ln 2+\dfrac{4}{3}\ln 3$
$\Rightarrow \left\{ \begin{aligned}
& a=\dfrac{-20}{3} \\
& b=\dfrac{4}{3} \\
& c=2 \\
\end{aligned} \right.\Rightarrow a+b+c=\dfrac{-10}{3}.$
Đổi cận $\left| \begin{aligned}
& x=0\Rightarrow t=1 \\
& x=1\Rightarrow t=2 \\
\end{aligned} \right.\Rightarrow I=\dfrac{2}{3}\int\limits_{1}^{2}{\dfrac{tdt}{{{t}^{2}}+5t+6}=\dfrac{2}{3}\int\limits_{1}^{2}{\dfrac{t}{\left( t+2 \right)\left( t+3 \right)}dt=\dfrac{2}{3}\int\limits_{1}^{2}{\dfrac{3\left( t+2 \right)-2\left( t+3 \right)}{\left( t+2 \right)\left( t+3 \right)}dt}}}$
$=\dfrac{2}{3}\int\limits_{1}^{2}{\left( \dfrac{3}{t+3}-\dfrac{2}{t+2} \right)dt=2\ln \left| t+3 \right|}\left| _{1}^{2} \right.-\dfrac{4}{3}\ln \left| t+2 \right|\left| _{1}^{2} \right.=2\ln \dfrac{5}{4}-\dfrac{4}{3}\ln \dfrac{4}{3}=2\ln 5-\dfrac{20}{3}\ln 2+\dfrac{4}{3}\ln 3$
$\Rightarrow \left\{ \begin{aligned}
& a=\dfrac{-20}{3} \\
& b=\dfrac{4}{3} \\
& c=2 \\
\end{aligned} \right.\Rightarrow a+b+c=\dfrac{-10}{3}.$
Đáp án A.