Google
Câu hỏi: Cho hàm số $f(x)$ liên tục và có đạo hàm trên $[2 ; 3]$ thỏa mãn $f^{\prime}(x)+\dfrac{2 x}{x^2-1} f(x)=x$ với $\forall x \in[2 ; 3]$, $f(2)=\dfrac{2}{3}$. Tính $I=\int_2^3 f(x) \mathrm{d} x$.
A. $\dfrac{4}{3}+\dfrac{1}{8} \ln \dfrac{3}{2}$.
B. $\dfrac{4}{3}+\dfrac{1}{8} \ln \dfrac{2}{3}$.
C. $\dfrac{53}{12}-\dfrac{1}{8} \ln 6$.
D. $\dfrac{53}{12}+\dfrac{1}{8} \ln \dfrac{3}{2}$.
A. $\dfrac{4}{3}+\dfrac{1}{8} \ln \dfrac{3}{2}$.
B. $\dfrac{4}{3}+\dfrac{1}{8} \ln \dfrac{2}{3}$.
C. $\dfrac{53}{12}-\dfrac{1}{8} \ln 6$.
D. $\dfrac{53}{12}+\dfrac{1}{8} \ln \dfrac{3}{2}$.
Ta có $f^{\prime}(x)+\dfrac{2 x}{x^2-1} f(x)=x(*)$
$
p(x)=\dfrac{2 x}{x^2-1} \Rightarrow P(x)=\int \dfrac{2 x}{x^2-1} \mathrm{~d} x=\int \dfrac{\mathrm{d}\left(x^2-1\right)}{x^2-1}=\ln \left|x^2-1\right|+C
$
Với $\forall x \in[2 ; 3] \Rightarrow x^2-1>0$, nhân hai vế của $(*)$ với $e^{\ln \left(x^2-1\right)}=x^2-1$, ta có:
$
\begin{aligned}
& \left(x^2-1\right) f^{\prime}(x)+2 x f(x)=\left(x^2-1\right) x \Leftrightarrow\left[\left(x^2-1\right) f(x)\right]^{\prime}=x^3-x \Leftrightarrow\left(x^2-1\right) f(x)=\dfrac{x^4}{4}- \\
& \dfrac{x^2}{2}+C \\
& \Rightarrow 3 f(2)=4-2+C \Leftrightarrow 2=2+C \Rightarrow C=0 . \\
& \text { Vậy } \Leftrightarrow\left(x^2-1\right) f(x)=\dfrac{x^4}{4}-\dfrac{x^2}{2} \Rightarrow f(x)=\dfrac{1}{4} \cdot \dfrac{x^4-2 x^2}{x^2-1}=\dfrac{1}{4}\left(x^2-1-\dfrac{1}{x^2-1}\right) . \\
& I=\int_2^3 \dfrac{1}{4}\left(x^2-1+\dfrac{1}{2}\left(\dfrac{1}{x+1}-\dfrac{1}{x-1}\right)\right) \mathrm{d} x=\left.\dfrac{1}{4}\left(\dfrac{x^3}{3}-x+\dfrac{1}{2} \ln \dfrac{x+1}{x-1}\right)\right|_2 ^3=\dfrac{4}{3}+\dfrac{1}{8} \ln \dfrac{2}{3} .
\end{aligned}
$
$
p(x)=\dfrac{2 x}{x^2-1} \Rightarrow P(x)=\int \dfrac{2 x}{x^2-1} \mathrm{~d} x=\int \dfrac{\mathrm{d}\left(x^2-1\right)}{x^2-1}=\ln \left|x^2-1\right|+C
$
Với $\forall x \in[2 ; 3] \Rightarrow x^2-1>0$, nhân hai vế của $(*)$ với $e^{\ln \left(x^2-1\right)}=x^2-1$, ta có:
$
\begin{aligned}
& \left(x^2-1\right) f^{\prime}(x)+2 x f(x)=\left(x^2-1\right) x \Leftrightarrow\left[\left(x^2-1\right) f(x)\right]^{\prime}=x^3-x \Leftrightarrow\left(x^2-1\right) f(x)=\dfrac{x^4}{4}- \\
& \dfrac{x^2}{2}+C \\
& \Rightarrow 3 f(2)=4-2+C \Leftrightarrow 2=2+C \Rightarrow C=0 . \\
& \text { Vậy } \Leftrightarrow\left(x^2-1\right) f(x)=\dfrac{x^4}{4}-\dfrac{x^2}{2} \Rightarrow f(x)=\dfrac{1}{4} \cdot \dfrac{x^4-2 x^2}{x^2-1}=\dfrac{1}{4}\left(x^2-1-\dfrac{1}{x^2-1}\right) . \\
& I=\int_2^3 \dfrac{1}{4}\left(x^2-1+\dfrac{1}{2}\left(\dfrac{1}{x+1}-\dfrac{1}{x-1}\right)\right) \mathrm{d} x=\left.\dfrac{1}{4}\left(\dfrac{x^3}{3}-x+\dfrac{1}{2} \ln \dfrac{x+1}{x-1}\right)\right|_2 ^3=\dfrac{4}{3}+\dfrac{1}{8} \ln \dfrac{2}{3} .
\end{aligned}
$
Đáp án B.