Contour integral ∫x²/(1+x²)³

Method. Let I = ∫_{−∞}^{∞} x²/(1 + x²)³ dx. Take f(z) = z²/(z² + 1)³ = z²/((z − i)³(z + i)³) on the closed contour C: segment [−R, R] plus upper semicircle Γ (R > 1).

Only z = i (pole of order 3) lies inside C.

Residue at z = i:

Res = (1/2!) d²/dz² [ (z − i)³ f(z) ] at z = i = (1/2) g″(i), where g(z) = z² (z + i)⁻³.

g′ = 2z (z + i)⁻³ − 3z² (z + i)⁻⁴

g″ = 2 (z + i)⁻³ − 12 z (z + i)⁻⁴ + 12 z² (z + i)⁻⁵

At z = i, z + i = 2i: (2i)⁻³ = i/8, (2i)⁻⁴ = 1/16, (2i)⁻⁵ = −i/32.

g″(i) = 2(i/8) − 12 i (1/16) + 12(−1)(−i/32) = i/4 − 3i/4 + 3i/8 = −i/8

Res = (1/2)(−i/8) = −i/16

Residue theorem:

∮_C f dz = 2πi (−i/16) = 2π/16 = π/8.

Arc integral: |f(z)| ≤ R²/(R² − 1)³ ~ 1/R⁴ on Γ, so the integral over Γ ≤ πR · 1/R⁴ → 0.

Answer: I = π/8

(Check with x = tan θ: I = ∫ sin²θ cos²θ dθ over (−π/2, π/2) = π/8.)