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Câu hỏi: Cho hàm số $f(x)$ thỏa mãn $f(1)=4$ và $\left(x^2+3\right)^2 f^{\prime}(x)=2 x \cdot f^2(x) ; f(x) \neq 0, \forall x \in \mathbb{R}$. Giá trị của $f(3)$ bằng
A. 12 .
B. 6 .
C. 2019 .
D. 9 .
A. 12 .
B. 6 .
C. 2019 .
D. 9 .
Vì $\left(x^2+3\right)^2 f^{\prime}(x)=2 x . f^2(x) ; f(x) \neq 0, \forall x \in \mathbb{R}$ nên
$
\begin{aligned}
& \Rightarrow \dfrac{f^{\prime}(x)}{f^2(x)}=\dfrac{2 x}{\left(x^2+3\right)^2} \Rightarrow \int_1^3 \dfrac{f^{\prime}(x)}{f^2(x)} \mathrm{d} x=\int_1^3 \dfrac{2 x}{\left(x^2+3\right)^2} \mathrm{~d} x \Leftrightarrow \int_1^3 \dfrac{\mathrm{d} f(x)}{f^2(x)}=\int_1^3 \dfrac{\mathrm{d}\left(x^2+3\right)}{\left(x^2+3\right)^2} \\
& \Leftrightarrow-\left.\dfrac{1}{f(x)}\right|_1 ^3=-\left.\dfrac{1}{x^2+3}\right|_1 ^3 \Leftrightarrow \dfrac{1}{f(1)}-\dfrac{1}{f(3)}=\dfrac{1}{4}-\dfrac{1}{12} \Leftrightarrow \dfrac{1}{4}-\dfrac{1}{f(3)}=\dfrac{1}{4}-\dfrac{1}{12} \\
& \Leftrightarrow f(3)=12 \text {. } \\
&
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow \dfrac{f^{\prime}(x)}{f^2(x)}=\dfrac{2 x}{\left(x^2+3\right)^2} \Rightarrow \int_1^3 \dfrac{f^{\prime}(x)}{f^2(x)} \mathrm{d} x=\int_1^3 \dfrac{2 x}{\left(x^2+3\right)^2} \mathrm{~d} x \Leftrightarrow \int_1^3 \dfrac{\mathrm{d} f(x)}{f^2(x)}=\int_1^3 \dfrac{\mathrm{d}\left(x^2+3\right)}{\left(x^2+3\right)^2} \\
& \Leftrightarrow-\left.\dfrac{1}{f(x)}\right|_1 ^3=-\left.\dfrac{1}{x^2+3}\right|_1 ^3 \Leftrightarrow \dfrac{1}{f(1)}-\dfrac{1}{f(3)}=\dfrac{1}{4}-\dfrac{1}{12} \Leftrightarrow \dfrac{1}{4}-\dfrac{1}{f(3)}=\dfrac{1}{4}-\dfrac{1}{12} \\
& \Leftrightarrow f(3)=12 \text {. } \\
&
\end{aligned}
$
Đáp án A.