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Câu hỏi: Cho hàm số $y=f\left( x \right)$ thỏa mãn điều kiện $\int\limits_{0}^{2}{\dfrac{{f}'\left( x \right)d\text{x}}{x+2}=3}$ và $f\left( 2 \right)-2f\left( 0 \right)=4$. Tính tích phân $I=\int\limits_{0}^{1}{\dfrac{f\left( 2\text{x} \right)d\text{x}}{{{\left( x+1 \right)}^{2}}}}$.
A. $I=-\dfrac{1}{2}$
B. $I=0$
C. $I=-2$
D. $I=4$
A. $I=-\dfrac{1}{2}$
B. $I=0$
C. $I=-2$
D. $I=4$
Đặt $\left\{ \begin{aligned}
& u=\dfrac{1}{x+2} \\
& dv={f}'\left( x \right)d\text{x} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{1}{{{\left( x+2 \right)}^{2}}} \\
& v=f\left( x \right) \\
\end{aligned} \right.$
Khi đó $\int\limits_{0}^{2}{\dfrac{{f}'\left( x \right)d\text{x}}{x+2}}=\left. \dfrac{f\left( x \right)}{x+2} \right|_{0}^{2}+\int\limits_{0}^{2}{\dfrac{f\left( x \right)d\text{x}}{{{\left( x+2 \right)}^{2}}}}=\dfrac{f\left( 2 \right)}{4}-\dfrac{f\left( 0 \right)}{2}+\int\limits_{0}^{2}{\dfrac{f\left( x \right)d\text{x}}{{{\left( x+2 \right)}^{2}}}}=1+\int\limits_{0}^{2}{\dfrac{f\left( x \right)d\text{x}}{{{\left( x+2 \right)}^{2}}}}$
Suy ra $K=\int\limits_{0}^{2}{\dfrac{f\left( x \right)d\text{x}}{{{\left( x+2 \right)}^{2}}}}=2\xrightarrow{x=2t}K=\int\limits_{0}^{1}{\dfrac{f\left( 2t \right)d2t}{{{\left( 2t+2 \right)}^{2}}}}=\int\limits_{0}^{1}{\dfrac{f\left( 2t \right)dt}{2{{\left( t+1 \right)}^{2}}}}=2$.
Vậy $\int\limits_{0}^{1}{\dfrac{f\left( 2t \right)dt}{{{\left( t+1 \right)}^{2}}}}=4$.
& u=\dfrac{1}{x+2} \\
& dv={f}'\left( x \right)d\text{x} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{1}{{{\left( x+2 \right)}^{2}}} \\
& v=f\left( x \right) \\
\end{aligned} \right.$
Khi đó $\int\limits_{0}^{2}{\dfrac{{f}'\left( x \right)d\text{x}}{x+2}}=\left. \dfrac{f\left( x \right)}{x+2} \right|_{0}^{2}+\int\limits_{0}^{2}{\dfrac{f\left( x \right)d\text{x}}{{{\left( x+2 \right)}^{2}}}}=\dfrac{f\left( 2 \right)}{4}-\dfrac{f\left( 0 \right)}{2}+\int\limits_{0}^{2}{\dfrac{f\left( x \right)d\text{x}}{{{\left( x+2 \right)}^{2}}}}=1+\int\limits_{0}^{2}{\dfrac{f\left( x \right)d\text{x}}{{{\left( x+2 \right)}^{2}}}}$
Suy ra $K=\int\limits_{0}^{2}{\dfrac{f\left( x \right)d\text{x}}{{{\left( x+2 \right)}^{2}}}}=2\xrightarrow{x=2t}K=\int\limits_{0}^{1}{\dfrac{f\left( 2t \right)d2t}{{{\left( 2t+2 \right)}^{2}}}}=\int\limits_{0}^{1}{\dfrac{f\left( 2t \right)dt}{2{{\left( t+1 \right)}^{2}}}}=2$.
Vậy $\int\limits_{0}^{1}{\dfrac{f\left( 2t \right)dt}{{{\left( t+1 \right)}^{2}}}}=4$.
Đáp án D.