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Câu hỏi: Cho hàm số $y=f(x)$ xác định và liên tục trên $\left[\dfrac{1}{2} ; 2\right]$, thỏa mãn $f(x)+f\left(\dfrac{1}{x}\right)=x^2+\dfrac{1}{x^2}+2$. Tính tích phân $I=\int_{\dfrac{1}{2}}^2 \dfrac{f(x)}{x^2+1} \mathrm{~d} x$.
A. $I=\dfrac{3}{2}$
B. $I=2$.
C. $I=\dfrac{5}{2}$.
D. $I=3$.
A. $I=\dfrac{3}{2}$
B. $I=2$.
C. $I=\dfrac{5}{2}$.
D. $I=3$.
Đặt $x=\dfrac{1}{t^{\prime}}$ suy ra $\mathrm{d} x=-\dfrac{1}{t^2} \mathrm{~d} t$. Đổi cận: $\left\{\begin{array}{l}x=\dfrac{1}{2} \rightarrow t=2 \\ x=2 \rightarrow t=\dfrac{1}{2}\end{array}\right.$.
Khi đó $I=\int_2^{\dfrac{1}{2}} \dfrac{f\left(\dfrac{1}{t}\right)}{t^2+1} \cdot\left(-\dfrac{1}{t^2}\right) \mathrm{d} t=\int_{\dfrac{1}{2}}^2 \dfrac{f\left(\dfrac{1}{t}\right)}{t^2+1} \mathrm{~d} t=\int_{\dfrac{1}{2}}^2 \dfrac{f\left(\dfrac{1}{x}\right)}{x^2+1} \mathrm{~d} x$.
Suy ra $2 I=\int_{\dfrac{1}{2}}^2 \dfrac{f(x)}{x^2+1} \mathrm{~d} x+\int_{\dfrac{1}{2}}^2 \dfrac{f\left(\dfrac{1}{x}\right)}{x^2+1} \mathrm{~d} x=\int_{\dfrac{1}{2}}^2 \dfrac{f(x)+f\left(\dfrac{1}{x}\right)}{x^2+1} \mathrm{~d} x=\int_{\dfrac{1}{2}}^2 \dfrac{x^2+\dfrac{1}{x^2}+2}{x^2+1} \mathrm{~d} x$
$
=\int_{\dfrac{1}{2}}^2 \dfrac{x^2+1}{x^2} \mathrm{~d} x=\int_{\dfrac{1}{2}}^2\left(1+\dfrac{1}{x^2}\right) \mathrm{d} x=\left.\left(x-\dfrac{1}{x}\right)\right|_{\dfrac{1}{2}} ^2=3 \rightarrow I=\dfrac{3}{2}
$
Khi đó $I=\int_2^{\dfrac{1}{2}} \dfrac{f\left(\dfrac{1}{t}\right)}{t^2+1} \cdot\left(-\dfrac{1}{t^2}\right) \mathrm{d} t=\int_{\dfrac{1}{2}}^2 \dfrac{f\left(\dfrac{1}{t}\right)}{t^2+1} \mathrm{~d} t=\int_{\dfrac{1}{2}}^2 \dfrac{f\left(\dfrac{1}{x}\right)}{x^2+1} \mathrm{~d} x$.
Suy ra $2 I=\int_{\dfrac{1}{2}}^2 \dfrac{f(x)}{x^2+1} \mathrm{~d} x+\int_{\dfrac{1}{2}}^2 \dfrac{f\left(\dfrac{1}{x}\right)}{x^2+1} \mathrm{~d} x=\int_{\dfrac{1}{2}}^2 \dfrac{f(x)+f\left(\dfrac{1}{x}\right)}{x^2+1} \mathrm{~d} x=\int_{\dfrac{1}{2}}^2 \dfrac{x^2+\dfrac{1}{x^2}+2}{x^2+1} \mathrm{~d} x$
$
=\int_{\dfrac{1}{2}}^2 \dfrac{x^2+1}{x^2} \mathrm{~d} x=\int_{\dfrac{1}{2}}^2\left(1+\dfrac{1}{x^2}\right) \mathrm{d} x=\left.\left(x-\dfrac{1}{x}\right)\right|_{\dfrac{1}{2}} ^2=3 \rightarrow I=\dfrac{3}{2}
$
Đáp án A.