Let's solve the integral using residue theory in **2 steps**, with an explanation after each step: --- ### **Step 1: Express the Integral Using Contour Integration** Given: \[ I = \int_0^\infty \frac{x^2}{(x^2+1)^3} \, dx \] Extend this to the whole real line (since the function is even): \[ I = \frac{1}{2} \int_{-\infty}^{\infty} \frac{x^2}{(x^2+1)^3} \, dx \] Now, consider the complex function: \[ f(z) = \frac{z^2}{(z^2+1)^3} \] We will integrate \( f(z) \) along a contour in the upper half-plane, which includes the real axis from \(-R\) to \(R\) and a semicircular arc of radius \(R\) in the upper half-plane. As \(R \rightarrow \infty\), the arc's contribution vanishes due to the degree of the denominator being higher than the numerator. **Explanation:** We have rewritten our real integral in terms of a contour integral over the real axis, which allows us to calculate it using the residue theorem. --- ### **Step 2: Apply the Residue Theorem** The poles of \( f(z) \) are at \( z^2 + 1 = 0 \implies z = i, -i \), both of order 3. Only the pole at \( z = i \) lies within the upper half-plane. We need the residue at \( z = i \): For a pole of order 3 at \( z_0 \): \[ \text{Res}(f, z_0) = \frac{1}{2!} \lim_{z \to z_0} \frac{d^2}{dz^2} \left[ (z-z_0)^3 f(z) \right] \] Here, \( (z-i)^3 f(z) = \frac{z^2}{(z+i)^3} \). Compute the second derivative at \( z=i \): Let \( g(z) = \frac{z^2}{(z+i)^3} \). First derivative: \[ g'(z) = \frac{2z(z+i)^3 - z^2 \cdot 3(z+i)^2}{(z+i)^6} = \frac{2z(z+i) - 3z^2}{(z+i)^4} \] Second derivative: \[ g''(z) = \frac{2(z+i) + 2z - 6z}{(z+i)^4} - \frac{4[2z(z+i) - 3z^2]}{(z+i)^5} \] But it's easier to use the general formula for residues at higher order poles. For a pole at \( z=i \): \[ \text{Res}(f, i) = \frac{1}{2!} \lim_{z \to i} \frac{d^2}{dz^2} \left( \frac{z^2}{(z+i)^3} \right) \] Let’s write \( h(z) = z^2 \) and \( k(z) = (z+i)^3 \). \[ \frac{d^2}{dz^2} \left( \frac{z^2}{(z+i)^3} \right) = \frac{d}{dz} \left( \frac{2z (z+i)^3 - 3z^2 (z+i)^2}{(z+i)^6} \right) \] Evaluate at \( z = i \): - \( z^2 = -1 \) - \( (z+i)^3 = (2i)^3 = 8i^3 = 8(-i) = -8i \) But let's use a more direct approach. For the even function: \[ \int_{-\infty}^{\infty} \frac{x^2}{(x^2+1)^3} dx = 2\pi i \cdot \text{Res}(f(z), z=i) \] So: \[ I = \frac{1}{2} \cdot 2\pi i \cdot \text{Res}(f(z), z=i) = \pi i \cdot \text{Res}(f(z), z=i) \] Now, compute the residue: Let \( z = i + \epsilon \), expand: \[ f(z) = \frac{z^2}{(z^2+1)^3} = \frac{z^2}{(z-i)^3 (z+i)^3} \] At \( z=i \), only the \((z-i)^3\) in the denominator is singular. Let’s use: \[ \text{Res}(f, i) = \frac{1}{2!} \lim_{z \to i} \frac{d^2}{dz^2} \left( (z-i)^3 \frac{z^2}{(z-i)^3 (z+i)^3} \right) = \frac{1}{2!} \lim_{z \to i} \frac{d^2}{dz^2} \left( \frac{z^2}{(z+i)^3} \right) \] Calculate derivatives: First derivative: \[ \frac{d}{dz} \left( \frac{z^2}{(z+i)^3} \right) = \frac{2z (z+i)^3 - z^2 \cdot 3 (z+i)^2}{(z+i)^6} = \frac{2z(z+i) - 3z^2}{(z+i)^4} \] Second derivative: \[ \frac{d}{dz} \left( \frac{2z(z+i) - 3z^2}{(z+i)^4} \right) = \frac{2(z+i) + 2z - 6z}{(z+i)^4} - \frac{4[2z(z+i) - 3z^2]}{(z+i)^5} \] Simplify \(2(z+i) + 2z - 6z = 2z + 2i + 2z - 6z = -2z + 2i\): So, \[ = \frac{-2z + 2i}{(z+i)^4} - \frac{8z(z+i) - 12z^2}{(z+i)^5} \] At \( z = i \): - \( z = i \) - \( (z+i) = 2i \) - \( (z+i)^4 = (2i)^4 = 16 \) - \( (z+i)^5 = (2i)^5 = 32i \) - \( z(z+i) = i \cdot 2i = 2i^2 = -2 \) - \( z^2 = (i)^2 = -1 \) So, First part: \[ \frac{-2z + 2i}{(z+i)^4} = \frac{-2i + 2i}{16} = 0 \] Second part: \[ - \frac{8z(z+i) - 12z^2}{(z+i)^5} = - \frac{8 \cdot (-2) - 12 \cdot (-1)}{32i} = - \frac{-16 + 12}{32i} = - \frac{-4}{32i} = \frac{1}{8i} \] Therefore, \[ \text{Res}(f, i) = \frac{1}{2} \cdot \frac{1}{8i} = \frac{1}{16i} \] So, \[ I = \pi i \cdot \frac{1}{16i} = \frac{\pi}{16} \] --- ## **Final Answer** \[ \boxed{\int_0^\infty \frac{x^2}{(x^2+1)^3} dx = \frac{\pi}{16}} \]