Residues of a function

Let I = ∫_{−∞}^{∞} x²/((x² + 1)(x² + 4)) dx. Consider f(z) = z²/((z² + 1)(z² + 4)) over a closed contour C: the real axis from −R to R and the upper semicircle Γ of radius R.

Poles of f: z = ±i, ±2i. In the upper half plane: z = i and z = 2i (simple poles).

Res at z = i:   z²/[(z + i)(z² + 4)] at z = i = (−1)/(2i · 3) = −1/(6i)
Res at z = 2i:  z²/[(z² + 1)(z + 2i)] at z = 2i = (−4)/((−3)(4i)) = 1/(3i)
Sum = −1/(6i) + 2/(6i) = 1/(6i)

On Γ, |f(z)| ≈ R²/R⁴ = 1/R², so |∫_Γ f dz| ≤ (1/R²)(πR) = π/R → 0 as R → ∞.

By the residue theorem:

I = 2πi × 1/(6i) = π/3

Answer: I = π/3. (This agrees with the formula π/(a + b) for ∫ x²/((x² + a²)(x² + b²)) dx with a = 1, b = 2.)