Cho hàm số \(y=f(x)\) xác định và liên tục trên \(...

A

- Hướng dẫn giải

Đặt \(x = \frac{1}{t},\) suy ra \({\rm{d}}x =  - \frac{1}{{{t^2}}}{\rm{d}}t.\) Đổi cận: \(\left\{ \begin{array}{l}
x = \frac{1}{2} \to t = 2\\
x = 2 \to t = \frac{1}{2}
\end{array} \right..\)

Khi đó \(I = \int\limits_2^{\frac{1}{2}} {\frac{{f\left( {\frac{1}{t}} \right)}}{{\frac{1}{{{t^2}}} + 1}}.\left( { - \frac{1}{{{t^2}}}} \right){\rm{d}}t}  = \int\limits_{\frac{1}{2}}^2 {\frac{{f\left( {\frac{1}{t}} \right)}}{{{t^2} + 1}}{\rm{d}}t}  = \int\limits_{\frac{1}{2}}^2 {\frac{{f\left( {\frac{1}{x}} \right)}}{{{x^2} + 1}}{\rm{d}}x} .\)

Suy ra \(2I = \int\limits_{\frac{1}{2}}^2 {\frac{{f\left( x \right)}}{{{x^2} + 1}}{\rm{d}}x}  + \int\limits_{\frac{1}{2}}^2 {\frac{{f\left( {\frac{1}{x}} \right)}}{{{x^2} + 1}}{\rm{d}}x}  = \int\limits_{\frac{1}{2}}^2 {\frac{{f\left( x \right) + f\left( {\frac{1}{x}} \right)}}{{{x^2} + 1}}{\rm{d}}x}  = \int\limits_{\frac{1}{2}}^2 {\frac{{{x^2} + \frac{1}{{{x^2}}} + 2}}{{{x^2} + 1}}{\rm{d}}x} \)

\( = \int\limits_{\frac{1}{2}}^2 {\frac{{{x^2} + 1}}{{{x^2}}}{\rm{d}}x}  = \int\limits_{\frac{1}{2}}^2 {\left( {1 + \frac{1}{{{x^2}}}} \right){\rm{d}}x}  = \left( {x - \frac{1}{x}} \right)\left| {\begin{array}{*{20}{c}}
2\\
{\frac{1}{2}}
\end{array}} \right. = 3 \to I = \frac{3}{2}.\)