Lời giải

Đặt \( t = x^3 \Rightarrow I = \frac{\pi}{12} \)

•   Bấm: \( I = 0.261799 = \frac{\pi}{12} = \frac{3.14159}{12}\)

\( J = \int_{0}^{1} \frac{x^4 - x^2 + 1 + x^2}{(x^2 + 1)(x^4 - x^2 + 1)} \, dx \)

    \( = \int_{0}^{1} \frac{1}{1 + x^2} \, dx + \int_{0}^{1} \frac{x^2}{1 + x^6} \, dx\)

    \( = \frac{\pi}{4} + \frac{\pi}{12} = \frac{\pi}{3}\)
 

Bấm: \( J = 1.047197 = \frac{\pi}{3} = \frac{3.14159}{3}\)