\(\displaystyle \int\limits_{ - \frac{1}{2}}^{\frac{1}{2}} {\frac{{x\left( {1 + {x^2} + {x^4}} \right)}}{{1 + {x^2}}}dx} \)\(\displaystyle = \int\limits_{ - \frac{1}{2}}^{\frac{1}{2}} {\left( {{x^3} + \frac{x}{{{x^2} + 1}}} \right)dx} \) \(\displaystyle = \int\limits_{ - \frac{1}{2}}^{\frac{1}{2}} {{x^3}dx} + \int\limits_{ - \frac{1}{2}}^{\frac{1}{2}} {\frac{{xdx}}{{{x^2} + 1}}dx} \) \(\displaystyle = I + J\)
Ta có: \(\displaystyle I = \int\limits_{ - \frac{1}{2}}^{\frac{1}{2}} {{x^3}dx} \)\(\displaystyle = \left. {\frac{{{x^4}}}{4}} \right|_{ - \frac{1}{2}}^{\frac{1}{2}} = \frac{1}{4}\left( {\frac{1}{{16}} - \frac{1}{{16}}} \right) = 0\)
Tính \(\displaystyle J = \int\limits_{ - \frac{1}{2}}^{\frac{1}{2}} {\frac{{xdx}}{{{x^2} + 1}}dx} \)\(\displaystyle = \frac{1}{2}\int\limits_{ - \frac{1}{2}}^{\frac{1}{2}} {\frac{{d\left( {{x^2} + 1} \right)}}{{{x^2} + 1}}} = \left. {\ln \left( {{x^2} + 1} \right)} \right|_{ - \frac{1}{2}}^{\frac{1}{2}} = 0\)
Vậy \(\displaystyle \int\limits_{ - \frac{1}{2}}^{\frac{1}{2}} {\frac{{x\left( {1 + {x^2} + {x^4}} \right)}}{{1 + {x^2}}}dx} = I + J = 0\).
Chọn A.
Chú ý:
Có thể chứng minh hàm số \(\displaystyle f\left( x \right) = \frac{{x\left( {1 + {x^2} + {x^4}} \right)}}{{1 + {x^2}}}\) là hàm số lẻ trên \(\displaystyle \left[ { - \frac{1}{2};\frac{1}{2}} \right]\) và sử dụng lý thuyết \(\displaystyle \int\limits_{ - a}^a {f\left( x \right)dx} = 0\) nếu hàm số \(\displaystyle f\left( x \right)\) lẻ trên \(\displaystyle \left[ { - a;a} \right]\).
