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Câu hỏi: Biết $\int\limits_{0}^{4}{\dfrac{\sqrt{2x+1}dx}{2x+3\sqrt{2x+1}+3}}=a+b\ln 2+c\ln \dfrac{5}{3}\left( a,b,c\in \mathbb{Z} \right)$. Tính $T=2a+b+c.$
A. $T=4.$
B. $T=2.$
C. $T=1.$
D. $T=3.$
A. $T=4.$
B. $T=2.$
C. $T=1.$
D. $T=3.$
$I=\int\limits_{0}^{4}{\dfrac{\sqrt{2x+1}dx}{2x+3\sqrt{2x+1}+3}}=\int\limits_{0}^{4}{\dfrac{\sqrt{2x+1}dx}{\left( \sqrt{2x+1}+1 \right)\left( \sqrt{2x+1}+2 \right)}}$
$=\int\limits_{0}^{4}{\dfrac{2\left( \sqrt{2x+1}+1 \right)-\left( \sqrt{2x+1}+2 \right)dx}{\left( \sqrt{2x+1}+1 \right)\left( \sqrt{2x+1}+2 \right)}}=\int\limits_{0}^{4}{\dfrac{2dx}{\left( \sqrt{2x+1}+2 \right)}}-\int\limits_{0}^{4}{\dfrac{dx}{\left( \sqrt{2x+1}+1 \right)}}.$
Đặt $u=\sqrt{2x+1}\Rightarrow u\text{d}u=dx.$
Với $x=0\Rightarrow u=1$, với $x=4\Rightarrow u=3.$
Suy ra $I=\int\limits_{1}^{.3}{\dfrac{2udu}{u+2}}-\int\limits_{1}^{.3}{\dfrac{udu}{u+1}}=\int\limits_{1}^{.3}{\left( 2-\dfrac{4}{u+2} \right)du}-\int\limits_{1}^{.3}{\left( 1-\dfrac{1}{u+1} \right)du}$
$=\left. \left( u-4\ln \left| u+2 \right|+\ln \left| u+1 \right| \right) \right|_{1}^{3}=2-4\ln \dfrac{5}{3}+\ln 2$
$\Rightarrow a=2,b=1,c=1\Rightarrow T=2.1+1-4=1.$
$=\int\limits_{0}^{4}{\dfrac{2\left( \sqrt{2x+1}+1 \right)-\left( \sqrt{2x+1}+2 \right)dx}{\left( \sqrt{2x+1}+1 \right)\left( \sqrt{2x+1}+2 \right)}}=\int\limits_{0}^{4}{\dfrac{2dx}{\left( \sqrt{2x+1}+2 \right)}}-\int\limits_{0}^{4}{\dfrac{dx}{\left( \sqrt{2x+1}+1 \right)}}.$
Đặt $u=\sqrt{2x+1}\Rightarrow u\text{d}u=dx.$
Với $x=0\Rightarrow u=1$, với $x=4\Rightarrow u=3.$
Suy ra $I=\int\limits_{1}^{.3}{\dfrac{2udu}{u+2}}-\int\limits_{1}^{.3}{\dfrac{udu}{u+1}}=\int\limits_{1}^{.3}{\left( 2-\dfrac{4}{u+2} \right)du}-\int\limits_{1}^{.3}{\left( 1-\dfrac{1}{u+1} \right)du}$
$=\left. \left( u-4\ln \left| u+2 \right|+\ln \left| u+1 \right| \right) \right|_{1}^{3}=2-4\ln \dfrac{5}{3}+\ln 2$
$\Rightarrow a=2,b=1,c=1\Rightarrow T=2.1+1-4=1.$
Đáp án C.