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Câu hỏi: Cho hàm số $f(x)=\left\{\begin{array}{cl}x^{2}+1 & \text { khi } x \geq 0 \\ 2 x^{2}+1 & \text { khi } x<0\end{array}\right.$. Tích phân $\int_{\dfrac{1}{\mathrm{e}}}^{\mathrm{e}} \dfrac{f^{\prime}(\ln x) \ln x}{x} \mathrm{~d} x$ bằng
A. $\dfrac{14}{3}$.
B. $-\dfrac{4}{3}$.
C. $-4$.
D. 2 .
A. $\dfrac{14}{3}$.
B. $-\dfrac{4}{3}$.
C. $-4$.
D. 2 .
Xét $I=\int_{\dfrac{1}{\mathrm{e}}}^{\mathrm{e}} \dfrac{f^{\prime}(\ln x) \ln x}{x} \mathrm{~d} x$
Đặt $\left\{\begin{array}{l}u=\ln x \\ \mathrm{~d} v=\dfrac{f^{\prime}(\ln x)}{x} \mathrm{~d} x\end{array} \Rightarrow\left\{\begin{array}{l}\mathrm{d} u=\dfrac{1}{x} \mathrm{~d} x \\ v=f(\ln x)\end{array} \Rightarrow I=\left.\ln x f(\ln x)\right|_{\dfrac{1}{\mathrm{e}}} ^{\mathrm{e}}-\int_{\dfrac{1}{\mathrm{e}}}^{\mathrm{e}} \dfrac{f(\ln x)}{x} \mathrm{~d} x\right.\right.$
$I=1 . f(1)+f(-1)-\int_{\dfrac{1}{\mathrm{e}}}^{\mathrm{e}} f(\ln x) \mathrm{d}(\ln x) \Leftrightarrow I=2+3-\int_{-1}^{1} f(t) \mathrm{d} t$
$\Leftrightarrow I=5-\int_{-1}^{0} f(t) \mathrm{d} t-\int_{0}^{1} f(t) \mathrm{d} t=5-\int_{-1}^{0}\left(2 x^{2}+1\right) \mathrm{d} x-\int_{0}^{1}\left(x^{2}+1\right) \mathrm{d} x=2 .$
Đặt $\left\{\begin{array}{l}u=\ln x \\ \mathrm{~d} v=\dfrac{f^{\prime}(\ln x)}{x} \mathrm{~d} x\end{array} \Rightarrow\left\{\begin{array}{l}\mathrm{d} u=\dfrac{1}{x} \mathrm{~d} x \\ v=f(\ln x)\end{array} \Rightarrow I=\left.\ln x f(\ln x)\right|_{\dfrac{1}{\mathrm{e}}} ^{\mathrm{e}}-\int_{\dfrac{1}{\mathrm{e}}}^{\mathrm{e}} \dfrac{f(\ln x)}{x} \mathrm{~d} x\right.\right.$
$I=1 . f(1)+f(-1)-\int_{\dfrac{1}{\mathrm{e}}}^{\mathrm{e}} f(\ln x) \mathrm{d}(\ln x) \Leftrightarrow I=2+3-\int_{-1}^{1} f(t) \mathrm{d} t$
$\Leftrightarrow I=5-\int_{-1}^{0} f(t) \mathrm{d} t-\int_{0}^{1} f(t) \mathrm{d} t=5-\int_{-1}^{0}\left(2 x^{2}+1\right) \mathrm{d} x-\int_{0}^{1}\left(x^{2}+1\right) \mathrm{d} x=2 .$
Đáp án D.