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Câu hỏi: Biết $\int\limits_{0}^{3}{x\ln \left( 2\text{x}+1 \right)d\text{x}}=\dfrac{35}{8}\ln a-\dfrac{b}{c}$, trong đó a, b, c là các số nguyên dương và $\dfrac{b}{c}$ là phân số tối giản. Tính $S=a-b+c$.
A. $S=-6$
B. $S=6$
C. $S=7$
D. $S=12$
A. $S=-6$
B. $S=6$
C. $S=7$
D. $S=12$
Ta có $I=\int\limits_{0}^{3}{x\ln \left( 2\text{x}+1 \right)d\text{x}}$. Đặt $\left\{ \begin{aligned}
& u=\ln \left( 2x+1 \right) \\
& dv=x\text{dx} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{2}{2\text{x}+1}d\text{x} \\
& v=\dfrac{{{x}^{2}}}{2} \\
\end{aligned} \right.$
$I=\int\limits_{0}^{3}{x\ln \left( 2\text{x}+1 \right)d\text{x}}=\left. \dfrac{{{x}^{2}}\ln \left( 2\text{x}+1 \right)}{2} \right|_{0}^{3}-\int\limits_{0}^{3}{\dfrac{{{x}^{2}}}{2\text{x}+1}d\text{x}}$
$=\dfrac{9}{2}\ln 7-\int\limits_{0}^{3}{\left( \dfrac{x}{2}-\dfrac{1}{4}+\dfrac{1}{4\left( 2\text{x}+1 \right)} \right)d\text{x}}=\dfrac{9}{2}\ln 7-\left. \left( \dfrac{{{x}^{2}}}{4}-\dfrac{1}{4}x+\dfrac{1}{8}\ln \left| 2x+1 \right| \right) \right|_{0}^{3}=\dfrac{35}{8}\ln 7-\dfrac{3}{2}$
$\Rightarrow \left\{ \begin{aligned}
& a=7 \\
& b=3 \\
& c=2 \\
\end{aligned} \right.\Rightarrow S=a-b+c=7-3+2=6$.
& u=\ln \left( 2x+1 \right) \\
& dv=x\text{dx} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{2}{2\text{x}+1}d\text{x} \\
& v=\dfrac{{{x}^{2}}}{2} \\
\end{aligned} \right.$
$I=\int\limits_{0}^{3}{x\ln \left( 2\text{x}+1 \right)d\text{x}}=\left. \dfrac{{{x}^{2}}\ln \left( 2\text{x}+1 \right)}{2} \right|_{0}^{3}-\int\limits_{0}^{3}{\dfrac{{{x}^{2}}}{2\text{x}+1}d\text{x}}$
$=\dfrac{9}{2}\ln 7-\int\limits_{0}^{3}{\left( \dfrac{x}{2}-\dfrac{1}{4}+\dfrac{1}{4\left( 2\text{x}+1 \right)} \right)d\text{x}}=\dfrac{9}{2}\ln 7-\left. \left( \dfrac{{{x}^{2}}}{4}-\dfrac{1}{4}x+\dfrac{1}{8}\ln \left| 2x+1 \right| \right) \right|_{0}^{3}=\dfrac{35}{8}\ln 7-\dfrac{3}{2}$
$\Rightarrow \left\{ \begin{aligned}
& a=7 \\
& b=3 \\
& c=2 \\
\end{aligned} \right.\Rightarrow S=a-b+c=7-3+2=6$.
Đáp án B.