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Câu hỏi: Cho $F\left( x \right)=\dfrac{1}{2{{x}^{2}}}$ là một nguyên hàm của hàm số $\dfrac{f\left( x \right)}{x}$ trên $\left( 0; +\infty \right)$. Tính tích phân $\int\limits_{1}^{2}{f\left( 2x+1 \right)\text{d}x}$.
A. $\int\limits_{1}^{2}{f\left( 2x+1 \right)\text{d}x}=\dfrac{2}{15}$.
B. $\int\limits_{1}^{2}{f\left( 2x+1 \right)\text{d}x}=-\dfrac{2}{15}$.
C. $\int\limits_{1}^{2}{f\left( 2x+1 \right)\text{d}x}=\dfrac{1}{15}$.
D. $\int\limits_{1}^{2}{f\left( 2x+1 \right)\text{d}x}=-\dfrac{1}{15}$.
A. $\int\limits_{1}^{2}{f\left( 2x+1 \right)\text{d}x}=\dfrac{2}{15}$.
B. $\int\limits_{1}^{2}{f\left( 2x+1 \right)\text{d}x}=-\dfrac{2}{15}$.
C. $\int\limits_{1}^{2}{f\left( 2x+1 \right)\text{d}x}=\dfrac{1}{15}$.
D. $\int\limits_{1}^{2}{f\left( 2x+1 \right)\text{d}x}=-\dfrac{1}{15}$.
Vì $F\left( x \right)$ là một nguyên hàm của hàm số $\dfrac{f\left( x \right)}{x}$ nên ${F}'\left( x \right)=\dfrac{f\left( x \right)}{x}$ $\Rightarrow \dfrac{f\left( x \right)}{x}={{\left( \dfrac{1}{2{{x}^{2}}} \right)}^{\prime }}=-\dfrac{1}{{{x}^{3}}}$.
$\Rightarrow $ $f\left( x \right)=-\dfrac{1}{{{x}^{2}}}$.
Do đó $\int\limits_{1}^{2}{f\left( 2x+1 \right)\text{d}x}=\int\limits_{1}^{2}{\dfrac{-1}{{{\left( 2x+1 \right)}^{2}}}\text{d}x}=\dfrac{1}{2\left( 2x+1 \right)}\left| \begin{aligned}
& 2 \\
& 1 \\
\end{aligned} \right.=-\dfrac{1}{15}$.
$\Rightarrow $ $f\left( x \right)=-\dfrac{1}{{{x}^{2}}}$.
Do đó $\int\limits_{1}^{2}{f\left( 2x+1 \right)\text{d}x}=\int\limits_{1}^{2}{\dfrac{-1}{{{\left( 2x+1 \right)}^{2}}}\text{d}x}=\dfrac{1}{2\left( 2x+1 \right)}\left| \begin{aligned}
& 2 \\
& 1 \\
\end{aligned} \right.=-\dfrac{1}{15}$.
Đáp án D.