∫(-1~1) 1/(x² + 4x + 5) dx= ∫(-1~1) 1/[(x + 2)² + 1]，公式∫ 1/(x²+a²) dx = (1/a)arctan(x/a)= arctan(x + 2) |(-1~1)= arctan(1 + 2) - arctan(2 - 1)= arctan(3) - arctan(1)= arctan(3) - π/4≈ 0.463648
