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Câu hỏi: Cho hàm số $f\left( x \right)=\left\{ \begin{array}{*{35}{l}}
{{x}^{2}}-2 & \text{ khi }x\ge 3 \\
2x+1 & \text{ khi }x<3 \\
\end{array} \right. $. Tích phân $ \int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{f\left( 3\tan x+1 \right)}{{{\cos }^{2}}x}~dx}$ bằng
A. $\dfrac{61}{3}$.
B. $\dfrac{61}{9}$.
C. $\dfrac{38}{3}$.
D. $\dfrac{38}{9}$.
{{x}^{2}}-2 & \text{ khi }x\ge 3 \\
2x+1 & \text{ khi }x<3 \\
\end{array} \right. $. Tích phân $ \int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{f\left( 3\tan x+1 \right)}{{{\cos }^{2}}x}~dx}$ bằng
A. $\dfrac{61}{3}$.
B. $\dfrac{61}{9}$.
C. $\dfrac{38}{3}$.
D. $\dfrac{38}{9}$.
Đặt $I=\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{f\left( 3\tan x+1 \right)}{{{\cos }^{2}}x}~dx}$
Đặt $u=3\tan x+1\Rightarrow du=3.\dfrac{dx}{{{\cos }^{2}}x}$ ;
Đổi cận $x=0\Rightarrow t=1;x=\dfrac{\pi }{4}\Rightarrow t=4.$
Do đó $I=\dfrac{1}{3}\int\limits_{1}^{4}{f\left( u \right)du=\dfrac{1}{3}\int\limits_{1}^{4}{f\left( x \right)dx}}=\dfrac{1}{3}\left[ \int\limits_{1}^{3}{\left( 2x+1 \right)du}+\int\limits_{3}^{4}{\left( {{x}^{2}}-2 \right)dx} \right]=\dfrac{1}{3}\left( 10+\dfrac{31}{3} \right)=\dfrac{61}{9}$
Đặt $u=3\tan x+1\Rightarrow du=3.\dfrac{dx}{{{\cos }^{2}}x}$ ;
Đổi cận $x=0\Rightarrow t=1;x=\dfrac{\pi }{4}\Rightarrow t=4.$
Do đó $I=\dfrac{1}{3}\int\limits_{1}^{4}{f\left( u \right)du=\dfrac{1}{3}\int\limits_{1}^{4}{f\left( x \right)dx}}=\dfrac{1}{3}\left[ \int\limits_{1}^{3}{\left( 2x+1 \right)du}+\int\limits_{3}^{4}{\left( {{x}^{2}}-2 \right)dx} \right]=\dfrac{1}{3}\left( 10+\dfrac{31}{3} \right)=\dfrac{61}{9}$
Đáp án B.