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Câu hỏi: Cho hàm số $f(x)$ có đạo hàm, liên tục trên đoạn $[1 ; 2]$ đồng thời thỏa mãn $f(2)=0$, $\int_1^2\left[f^{\prime}(x)\right]^2 \mathrm{~d} x=\dfrac{5}{12}+\ln \dfrac{2}{3}$ và $\int_1^2 \dfrac{f(x)}{(x+1)^2} \mathrm{~d} x=-\dfrac{5}{12}+\ln \dfrac{3}{2}$. Tính $I=\int_1^2 f(x) \mathrm{d} x$.
A. $I=\dfrac{3}{4}+2 \ln \dfrac{2}{3}$.
B. $I=\ln \dfrac{2}{3}$.
C. $I=\dfrac{3}{4}+2 \ln \dfrac{3}{2}$.
D. $I=\dfrac{3}{4}+2 \ln \dfrac{2}{3}$.
A. $I=\dfrac{3}{4}+2 \ln \dfrac{2}{3}$.
B. $I=\ln \dfrac{2}{3}$.
C. $I=\dfrac{3}{4}+2 \ln \dfrac{3}{2}$.
D. $I=\dfrac{3}{4}+2 \ln \dfrac{2}{3}$.
$
+ \text { Đặt }\left\{\begin{array} { l }
{ u = f ( x ) } \\
{ \mathrm { d } v = \dfrac { 1 } { ( x + 1 ) ^ { 2 } } \mathrm { d } x }
\end{array} \Rightarrow \left\{\begin{array}{l}
\mathrm{d} u=f^{\prime}(x) \mathrm{d} x \\
v=\dfrac{1}{2}\left(\dfrac{x-1}{x+1}\right)
\end{array} .\right.\right.
$
Khi đó
$
\begin{aligned}
& \int_1^2 \dfrac{f(x)}{(x+1)^2} \mathrm{~d} x=\dfrac{1}{2}\left[\left.f(x) \dfrac{x-1}{x+1}\right|_1 ^2-\int_1^2 \dfrac{x-1}{x+1} f^{\prime}(x) \mathrm{d} x\right] \\
& \Leftrightarrow-\dfrac{5}{12}+\ln \dfrac{3}{2}=\dfrac{1}{2}\left[f(2) \dfrac{1}{3}-\int_1^2 \dfrac{x-1}{\mathrm{x}+1} f^{\prime}(x) \mathrm{d} x\right] \\
& \Leftrightarrow \dfrac{5}{6}-2 \ln \dfrac{3}{2}=\int_1^2 \dfrac{x-1}{x+1} f^{\prime}(x) \mathrm{d} x \quad(1) . \\
& \text { Xét } \int_1^2\left(\dfrac{x-1}{x+1}\right)^2 \mathrm{~d} x=\int_1^2\left(1-\dfrac{2}{x+1}\right)^2 \mathrm{~d} x \\
& =\int_1^2\left(1-\dfrac{4}{x+1}+\dfrac{4}{(x+1)^2}\right) \mathrm{d} x=\left.\left(x-4 \ln |x+1|-\dfrac{4}{x+1}\right)\right|_1 ^2=1-4 \ln 3+4 \ln 2-\dfrac{4}{3}+2=\dfrac{5}{3}-4 \ln \dfrac{3}{2} \\
& \Rightarrow \dfrac{1}{4} \int_1^2\left(\dfrac{x-1}{x+1}\right)^2 \mathrm{~d} x=\dfrac{5}{12}-\ln \dfrac{3}{2} \quad \text { (2). }
\end{aligned}
$
Theo đề $\int_1^2\left[f^{\prime}(x)\right]^2 \mathrm{~d} x=\dfrac{5}{12}-\ln \dfrac{3}{2}$ (3)
Từ (1), (2), (3) ta có
$
\begin{aligned}
& \int_1^2\left[\dfrac{1}{2}\left(\dfrac{x-1}{x+1}\right)-f^{\prime}(x)\right]^2 \mathrm{~d} x=0 \Rightarrow \dfrac{1}{2}\left(\dfrac{x-1}{x+1}\right)-f^{\prime}(x)=0 \Leftrightarrow f^{\prime}(x)=\dfrac{1}{2}\left(\dfrac{x-1}{x+1}\right) . \\
& \Rightarrow f(x)=\dfrac{1}{2}[x-2 \ln (x+1)]+C \\
& \Rightarrow f(2)=\dfrac{1}{2}[2-2 \ln 3]+C=0 \Rightarrow C=\ln 3-1 \\
& \Rightarrow f(\mathrm{x})=\dfrac{1}{2}[x-2 \ln (x+1)]+\ln 3-1 \\
& I=\int_1^2\left(\dfrac{1}{2}[x-2 \ln (x+1)]+\ln 3-1\right) \mathrm{d} x \\
& =\left.\left[\dfrac{x^2}{4}+(\ln 3-1) x\right]\right|_1 ^2-\int_1^2 \ln (x+1) \mathrm{d} x=-\dfrac{1}{4}+\ln 3-\int_1^2 \ln (x+1) \mathrm{d} x \\
& =-\dfrac{1}{4}+\ln 3-\left[\left.(x+1) \ln (x+1)\right|_1 ^2-\int_1^2 x \mathrm{~d} x\right] \\
& =-\dfrac{1}{4}+\ln 3-[3 \ln 3-2 \ln 2-1]=\dfrac{3}{2}-2 \ln \dfrac{2}{3} .
\end{aligned}
$
+ \text { Đặt }\left\{\begin{array} { l }
{ u = f ( x ) } \\
{ \mathrm { d } v = \dfrac { 1 } { ( x + 1 ) ^ { 2 } } \mathrm { d } x }
\end{array} \Rightarrow \left\{\begin{array}{l}
\mathrm{d} u=f^{\prime}(x) \mathrm{d} x \\
v=\dfrac{1}{2}\left(\dfrac{x-1}{x+1}\right)
\end{array} .\right.\right.
$
Khi đó
$
\begin{aligned}
& \int_1^2 \dfrac{f(x)}{(x+1)^2} \mathrm{~d} x=\dfrac{1}{2}\left[\left.f(x) \dfrac{x-1}{x+1}\right|_1 ^2-\int_1^2 \dfrac{x-1}{x+1} f^{\prime}(x) \mathrm{d} x\right] \\
& \Leftrightarrow-\dfrac{5}{12}+\ln \dfrac{3}{2}=\dfrac{1}{2}\left[f(2) \dfrac{1}{3}-\int_1^2 \dfrac{x-1}{\mathrm{x}+1} f^{\prime}(x) \mathrm{d} x\right] \\
& \Leftrightarrow \dfrac{5}{6}-2 \ln \dfrac{3}{2}=\int_1^2 \dfrac{x-1}{x+1} f^{\prime}(x) \mathrm{d} x \quad(1) . \\
& \text { Xét } \int_1^2\left(\dfrac{x-1}{x+1}\right)^2 \mathrm{~d} x=\int_1^2\left(1-\dfrac{2}{x+1}\right)^2 \mathrm{~d} x \\
& =\int_1^2\left(1-\dfrac{4}{x+1}+\dfrac{4}{(x+1)^2}\right) \mathrm{d} x=\left.\left(x-4 \ln |x+1|-\dfrac{4}{x+1}\right)\right|_1 ^2=1-4 \ln 3+4 \ln 2-\dfrac{4}{3}+2=\dfrac{5}{3}-4 \ln \dfrac{3}{2} \\
& \Rightarrow \dfrac{1}{4} \int_1^2\left(\dfrac{x-1}{x+1}\right)^2 \mathrm{~d} x=\dfrac{5}{12}-\ln \dfrac{3}{2} \quad \text { (2). }
\end{aligned}
$
Theo đề $\int_1^2\left[f^{\prime}(x)\right]^2 \mathrm{~d} x=\dfrac{5}{12}-\ln \dfrac{3}{2}$ (3)
Từ (1), (2), (3) ta có
$
\begin{aligned}
& \int_1^2\left[\dfrac{1}{2}\left(\dfrac{x-1}{x+1}\right)-f^{\prime}(x)\right]^2 \mathrm{~d} x=0 \Rightarrow \dfrac{1}{2}\left(\dfrac{x-1}{x+1}\right)-f^{\prime}(x)=0 \Leftrightarrow f^{\prime}(x)=\dfrac{1}{2}\left(\dfrac{x-1}{x+1}\right) . \\
& \Rightarrow f(x)=\dfrac{1}{2}[x-2 \ln (x+1)]+C \\
& \Rightarrow f(2)=\dfrac{1}{2}[2-2 \ln 3]+C=0 \Rightarrow C=\ln 3-1 \\
& \Rightarrow f(\mathrm{x})=\dfrac{1}{2}[x-2 \ln (x+1)]+\ln 3-1 \\
& I=\int_1^2\left(\dfrac{1}{2}[x-2 \ln (x+1)]+\ln 3-1\right) \mathrm{d} x \\
& =\left.\left[\dfrac{x^2}{4}+(\ln 3-1) x\right]\right|_1 ^2-\int_1^2 \ln (x+1) \mathrm{d} x=-\dfrac{1}{4}+\ln 3-\int_1^2 \ln (x+1) \mathrm{d} x \\
& =-\dfrac{1}{4}+\ln 3-\left[\left.(x+1) \ln (x+1)\right|_1 ^2-\int_1^2 x \mathrm{~d} x\right] \\
& =-\dfrac{1}{4}+\ln 3-[3 \ln 3-2 \ln 2-1]=\dfrac{3}{2}-2 \ln \dfrac{2}{3} .
\end{aligned}
$
Đáp án A.