91Ó°ÊÓ

With a given integral, we can perform partial fraction decomposition before computing the integral. Use Ansatz to find proper decomposition (A, B, and C are constants to be determined): $$\frac{z+1}{z(z^2+4)} = \frac{A}{z} + \frac{Bz + C}{z^2 + 4}$$ Now, clear the fractions by multiplying both sides by \(z(z^2+4)\): $$(z+1) = A(z^2 + 4) + (Bz + C)z$$

Expand the expressions on the right side and collect like terms: $$z + 1 = (A + B)z^2 + (C)z + 4A$$ By comparing the coefficients of the powers of z, we get the following system of linear equations: $$A + B = 0$$ $$C = 1$$ $$4A = 1$$ Solve this system to obtain \(A=\frac{1}{4}\), \(B=-\frac{1}{4}\), and \(C=1\). Now we can rewrite the original integrand as follows: $$\frac{z+1}{z(z^2+4)} = \frac{\frac{1}{4}}{z} + \frac{-\frac{1}{4}z + 1}{z^2 + 4}$$

Integrate the decomposed function term by term with respect to z: $$\int \frac{z+1}{z(z^2+4)} dz = \int \left(\frac{\frac{1}{4}}{z} + \frac{-\frac{1}{4}z + 1}{z^2 + 4}\right) dz$$ $$= \frac{1}{4} \int \frac{1}{z} dz - \frac{1}{4}\int \frac{z}{z^2 + 4}dz + \int \frac{1}{z^2+4}dz$$

Evaluate the integrals term by term: 1. \(\frac{1}{4} \int \frac{1}{z} dz =\frac{1}{4}\ln (|z|) + C_1\) 2. \(-\frac{1}{4}\int \frac{z}{z^2 + 4}dz = -\frac{1}{8} \ln (z^2+4)+ C_2\) 3. Integrate \(\int \frac{1}{z^2+4}dz\) using the substitution method. Let \(u=z\) and \(dv=\frac{1}{z^2+4}dz\), then \(du=dz\) and \(v=\frac{1}{2}\arctan(\frac{z}{2})\). So, using the integration by parts formula, we have: $$\int \frac{1}{z^2+4}dz = \frac{1}{2} z\arctan(\frac{z}{2}) - \frac{1}{2} \int z\frac{1}{4+z^2}d(\frac{z}{2})$$ $$= \frac{1}{2} z\arctan(\frac{z}{2}) - \frac{1}{4} \int \arctan(\frac{z}{2})dz$$ Now, use integration by parts again with \(w=\arctan(\frac{z}{2})\) and \(dx=dz\), so \(dw=\frac{1}{4+z^2}dz\) and \(x=z\). Apply the integration by parts formula: $$\int \arctan(\frac{z}{2})dz = z\arctan(\frac{z}{2}) - \int z\frac{1}{4+z^2}dz$$ $$ = z\arctan(\frac{z}{2}) -\frac{1}{2} \ln (z^2+4) + C_3$$ Finally, combining all three integrals: $$\int \frac{z+1}{z(z^2+4)} dz = \frac{1}{4}\ln (|z|) -\frac{1}{8} \ln (z^2+4) + \frac{1}{2} z\arctan(\frac{z}{2}) - \frac{1}{4}z\arctan(\frac{z}{2})+\frac{1}{8}\ln (z^2+4) + C$$ Simplify the expression: $$\int \frac{z+1}{z(z^2+4)} dz = \frac{1}{4}\ln (|z|) + \frac{1}{4} z\arctan(\frac{z}{2}) + C$$