Bài tập 3.56 trang 183 SBT Toán 12

Bài tập 3.56 trang 183 SBT Toán 12

\(\begin{array}{l}
\int \limits_{ - \frac{1}{2}}^{\frac{1}{2}} \frac{{x\left( {1 + {x^2} + {x^4}} \right)}}{{1 + {x^2}}}dx = \int \limits_{ - \frac{1}{2}}^{\frac{1}{2}} \left( {{x^3} + \frac{x}{{{x^2} + 1}}} \right)dx\\
 = \int \limits_{ - \frac{1}{2}}^{\frac{1}{2}} {x^3}dx + \int \limits_{ - \frac{1}{2}}^{\frac{1}{2}} \frac{{xdx}}{{{x^2} + 1}}dx = I + J
\end{array}\)

Ta có: \(I = \int \limits_{ - \frac{1}{2}}^{\frac{1}{2}} {x^3}dx = \left. {\frac{{{x^4}}}{4}} \right|_{ - \frac{1}{2}}^{\frac{1}{2}}\)

\( = \frac{1}{4}\left( {\frac{1}{{16}} - \frac{1}{{16}}} \right) = 0\)

Tính \(J = \int \limits_{ - \frac{1}{2}}^{\frac{1}{2}} \frac{{xdx}}{{{x^2} + 1}}dx\)

\( = \frac{1}{2}\int\limits_{ - \frac{1}{2}}^{\frac{1}{2}} {\frac{{d\left( {{x^2} + 1} \right)}}{{{x^2} + 1}} = {\mathop{\rm l}\nolimits} \left. {n\left( {{x^2} + 1} \right)} \right|_{ - \frac{1}{2}}^{\frac{1}{2}} = 0} \)

Vậy \(\int \limits_{ - \frac{1}{2}}^{\frac{1}{2}} \frac{{x\left( {1 + {x^2} + {x^4}} \right)}}{{1 + {x^2}}}dx = I + J = 0\)

Chọn A.

-- Mod Toán 12