Google
Câu hỏi: CHo $I=\int\limits_{1}^{2}{\dfrac{x+\ln x}{{{\left( x+1 \right)}^{2}}}dx=\dfrac{a}{b}\ln 2}-\dfrac{1}{c}$ với $a,b,c$ là các số nguyên dương và các phân số 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 $\left\{ \begin{aligned}
& x+\ln x=u \\
& \dfrac{1}{{{\left( x+1 \right)}^{2}}}dx=dv \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \dfrac{1+x}{x}dx=du \\
& -\dfrac{1}{x+1}=v \\
\end{aligned} \right.$
Khi đó $I=\int\limits_{1}^{2}{\dfrac{x+\ln x}{{{\left( x+1 \right)}^{2}}}dx=-\dfrac{1}{x+1}\left( x+\ln x \right)\left| _{\begin{smallmatrix}
\\
1
\end{smallmatrix}}^{\begin{smallmatrix}
2 \\
\end{smallmatrix}} \right.+\int\limits_{1}^{2}{\dfrac{1+x}{x}.\dfrac{1}{x+1}dx=-\dfrac{1}{3}\left( 2+\ln 2 \right)+\dfrac{1}{2}}+\int\limits_{1}^{2}{\dfrac{1}{x}dx}}$
$=-\dfrac{1}{3}\left( 2+\ln 2 \right)+\dfrac{1}{2}+\ln \left| x \right|\left| _{1}^{2} \right.=\dfrac{2}{3}\ln 2-\dfrac{1}{6}$
Vậy $a=2;b=3;c=6\Rightarrow S=\dfrac{a+b}{c}=\dfrac{5}{6}.$
& x+\ln x=u \\
& \dfrac{1}{{{\left( x+1 \right)}^{2}}}dx=dv \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \dfrac{1+x}{x}dx=du \\
& -\dfrac{1}{x+1}=v \\
\end{aligned} \right.$
Khi đó $I=\int\limits_{1}^{2}{\dfrac{x+\ln x}{{{\left( x+1 \right)}^{2}}}dx=-\dfrac{1}{x+1}\left( x+\ln x \right)\left| _{\begin{smallmatrix}
\\
1
\end{smallmatrix}}^{\begin{smallmatrix}
2 \\
\end{smallmatrix}} \right.+\int\limits_{1}^{2}{\dfrac{1+x}{x}.\dfrac{1}{x+1}dx=-\dfrac{1}{3}\left( 2+\ln 2 \right)+\dfrac{1}{2}}+\int\limits_{1}^{2}{\dfrac{1}{x}dx}}$
$=-\dfrac{1}{3}\left( 2+\ln 2 \right)+\dfrac{1}{2}+\ln \left| x \right|\left| _{1}^{2} \right.=\dfrac{2}{3}\ln 2-\dfrac{1}{6}$
Vậy $a=2;b=3;c=6\Rightarrow S=\dfrac{a+b}{c}=\dfrac{5}{6}.$
Đáp án B.