Contour integral ∫x²/((1+x²)(x²+4))

Method. Let I = ∫_{−∞}^{∞} x²/((1 + x²)(x² + 4)) dx. Consider f(z) = z²/((z² + 1)(z² + 4)) around the closed contour C made of the real segment [−R, R] and the upper semicircle Γ: |z| = R.

Poles of f: z = ±i and z = ±2i (all simple). Those in the upper half-plane (inside C for R > 2): z = i and z = 2i.

Residues:

• At z = i: Res = lim (z − i) f(z) = i²/((2i)(−1 + 4)) = −1/(6i)
• At z = 2i: Res = (2i)²/(((2i)² + 1)(4i)) = −4/((−3)(4i)) = 1/(3i)

Sum = −1/(6i) + 1/(3i) = 1/(6i).

Residue theorem: ∮_C f dz = 2πi · 1/(6i) = π/3.

Semicircle vanishes: on Γ, |f(z)| ≤ R²/((R² − 1)(R² − 4)) ~ 1/R², so |∫_Γ f dz| ≤ (π R)(1/R²) → 0 as R → ∞.

Hence ∫_{−∞}^{∞} f(x) dx = π/3.

Answer: I = π/3

(Check by partial fractions: x²/((x² + 1)(x² + 4)) = (−1/3)/(x² + 1) + (4/3)/(x² + 4); integral = −π/3 + (4/3)(π/2) = π/3.)