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Câu hỏi: Biết $\int\limits_{0}^{3}{\dfrac{x}{4+2\sqrt{x+1}}dx}=\dfrac{a}{3}+b\ln 2+c\ln 3,$ trong đó a, b, c là các số nguyên. Tính $T=a+b+c.$
A. $T=1.$
B. $T=4.$
C. $T=3.$
D. $T=6.$
A. $T=1.$
B. $T=4.$
C. $T=3.$
D. $T=6.$
Đặt $\sqrt{x+1}=t\Rightarrow x={{t}^{2}}-1\Rightarrow dx=2tdt$
Đổi cận: $x=0\Rightarrow t=1;x=3\Rightarrow t=2$
$\int\limits_{0}^{3}{\dfrac{x}{4+2\sqrt{x+1}}dx}=\int\limits_{1}^{2}{\dfrac{{{t}^{2}}-1}{4+2t}.2tdt}$
$=\int\limits_{1}^{2}{\dfrac{{{t}^{3}}-1}{t+2}dt}=\int\limits_{1}^{2}{\left( {{t}^{2}}-2t+3-\dfrac{6}{t+2} \right)dt}=\left. \left( \dfrac{1}{3}{{t}^{3}}-{{t}^{2}}+3t-6\ln \left| t+2 \right| \right) \right|_{1}^{2}$
$\begin{aligned}
& =\left( \dfrac{14}{3}-12\ln 2 \right)-\left( \dfrac{7}{3}-6\ln 3 \right)=\dfrac{7}{3}-12\ln 2+6\ln 3 \\
& \Rightarrow a=7;b=-12;c=6\Rightarrow T=a+b+c=1 \\
\end{aligned}$
Đổi cận: $x=0\Rightarrow t=1;x=3\Rightarrow t=2$
$\int\limits_{0}^{3}{\dfrac{x}{4+2\sqrt{x+1}}dx}=\int\limits_{1}^{2}{\dfrac{{{t}^{2}}-1}{4+2t}.2tdt}$
$=\int\limits_{1}^{2}{\dfrac{{{t}^{3}}-1}{t+2}dt}=\int\limits_{1}^{2}{\left( {{t}^{2}}-2t+3-\dfrac{6}{t+2} \right)dt}=\left. \left( \dfrac{1}{3}{{t}^{3}}-{{t}^{2}}+3t-6\ln \left| t+2 \right| \right) \right|_{1}^{2}$
$\begin{aligned}
& =\left( \dfrac{14}{3}-12\ln 2 \right)-\left( \dfrac{7}{3}-6\ln 3 \right)=\dfrac{7}{3}-12\ln 2+6\ln 3 \\
& \Rightarrow a=7;b=-12;c=6\Rightarrow T=a+b+c=1 \\
\end{aligned}$
Đáp án A.