1. **Stating the problem:** We want to decompose the function $$y(p) = \frac{1}{p^2 + 4} + \frac{1 - e^{-p}}{p(p^2 + 4)}$$ into partial fractions. 2. **Rewrite the expression:** Combine the terms over the common denominator $$p(p^2 + 4)$$: $$y(p) = \frac{p}{p(p^2 + 4)} + \frac{1 - e^{-p}}{p(p^2 + 4)} = \frac{p + 1 - e^{-p}}{p(p^2 + 4)}$$ 3. **Set up partial fractions:** Since the denominator factors as $$p(p^2 + 4)$$, and $$p^2 + 4$$ is irreducible over the reals, the decomposition is: $$\frac{p + 1 - e^{-p}}{p(p^2 + 4)} = \frac{A}{p} + \frac{Bp + C}{p^2 + 4}$$ 4. **Multiply both sides by the denominator:** $$p + 1 - e^{-p} = A(p^2 + 4) + (Bp + C)p = A p^2 + 4A + B p^2 + C p$$ 5. **Group like terms:** $$p + 1 - e^{-p} = (A + B) p^2 + C p + 4 A$$ 6. **Equate coefficients:** - Coefficient of $$p^2$$: $$0 = A + B$$ - Coefficient of $$p$$: $$1 = C$$ - Constant term: $$1 - e^{-p} = 4 A$$ 7. **Solve for constants:** - From constant term: $$4 A = 1 - e^{-p} \implies A = \frac{1 - e^{-p}}{4}$$ - From $$p^2$$ term: $$B = -A = -\frac{1 - e^{-p}}{4}$$ - From $$p$$ term: $$C = 1$$ 8. **Final partial fraction decomposition:** $$y(p) = \frac{\frac{1 - e^{-p}}{4}}{p} + \frac{-\frac{1 - e^{-p}}{4} p + 1}{p^2 + 4}$$ This expresses the original function as a sum of simpler rational functions suitable for further analysis or inverse transforms.