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Câu hỏi: Cho $I=\underset{1}{\overset{2}{\mathop \int }} \dfrac{x+\ln x}{{{\left( x+1 \right)}^{2}}}dx=\dfrac{a}{b}\ln 2-\dfrac{1}{c}$ với $a,b,c$ là các số nguyên dương và $\dfrac{a}{b}$ là phân số tối giản. Tính giá trị của biểu thức $S=\dfrac{a+b}{c}.$
A. $S=\dfrac{2}{3}.$
B. $S=\dfrac{5}{6}.$
C. $S=\dfrac{1}{2}.$
D. $S=\dfrac{1}{3}.$
A. $S=\dfrac{2}{3}.$
B. $S=\dfrac{5}{6}.$
C. $S=\dfrac{1}{2}.$
D. $S=\dfrac{1}{3}.$
Đặt $\left\{ \begin{array}{*{35}{l}}
u=x+\ln x \\
dv=\dfrac{dx}{{{\left( x+1 \right)}^{2}}} \\
\end{array} \right.\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
du=\dfrac{x+1}{x}dx \\
v=-\dfrac{1}{x+1} \\
\end{array} \right. $ $ \Rightarrow I=\left. -\dfrac{x+\ln x}{x+1} \right|_{1}^{2}+\int\limits_{1}^{2}{\dfrac{1}{x}dx}=\left. \left( \ln x-\dfrac{x+\ln x}{x+1} \right) \right|_{1}^{2}$
$=\left( \ln 2-\dfrac{2+\ln 2}{3} \right)-\left( \ln 1-\dfrac{1+\ln 1}{2} \right)=\dfrac{2}{3}\ln 2-\dfrac{1}{6}=\dfrac{a}{b}\ln 2-\dfrac{1}{c}\Rightarrow \left\{ \begin{array}{*{35}{l}}
a=2 \$/B]
b=3 \$/B]
c=6 \$/B]
\end{array} \right..$
Vậy $S=\frac{a+b}{c}=\frac{5}{6}.$ Chọn B
u=x+\ln x \\
dv=\dfrac{dx}{{{\left( x+1 \right)}^{2}}} \\
\end{array} \right.\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
du=\dfrac{x+1}{x}dx \\
v=-\dfrac{1}{x+1} \\
\end{array} \right. $ $ \Rightarrow I=\left. -\dfrac{x+\ln x}{x+1} \right|_{1}^{2}+\int\limits_{1}^{2}{\dfrac{1}{x}dx}=\left. \left( \ln x-\dfrac{x+\ln x}{x+1} \right) \right|_{1}^{2}$
$=\left( \ln 2-\dfrac{2+\ln 2}{3} \right)-\left( \ln 1-\dfrac{1+\ln 1}{2} \right)=\dfrac{2}{3}\ln 2-\dfrac{1}{6}=\dfrac{a}{b}\ln 2-\dfrac{1}{c}\Rightarrow \left\{ \begin{array}{*{35}{l}}
a=2 \$/B]
b=3 \$/B]
c=6 \$/B]
\end{array} \right..$
Vậy $S=\frac{a+b}{c}=\frac{5}{6}.$ Chọn B
Đáp án B.