Answer
$$\frac{1}{4}\ln \left| z \right| - \frac{1}{4}\ln \left( {{z^2} + 4} \right) + {\tan ^{ - 1}}\left( z \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{z + 1}}{{z\left( {{z^2} + 4} \right)}}dz} \cr
& {\text{partial fractions}} \cr
& \frac{{z + 1}}{{z\left( {{z^2} + 4} \right)}} = \frac{A}{z} + \frac{{Bz + C}}{{{z^2} + 4}} \cr
& {\text{multiplying}} \cr
& z + 1 = A\left( {{z^2} + 4} \right) + \left( {Bz + C} \right)z \cr
& z + 1 = A{z^2} + 4A + B{z^2} + Cz \cr
& z + 1 = \left( {A{z^2} + B{z^2}} \right) + Cz + 4A \cr
& {\text{by equating the coefficients}} \cr
& {z^2}:{\text{ }}A + B = 0 \cr
& z:{\text{ }}Cz = 1 \cr
& {z^0}:{\text{ }}4A = 1 \cr
& {\text{Solving these equations}} \cr
& A = 1/4 \cr
& B = - 1/4 \cr
& C = 1 \cr
& {\text{substituting constants}} \cr
& \frac{A}{z} + \frac{{Bz + C}}{{{z^2} + 4}} = \frac{{1/4}}{z} + \frac{{ - \left( {1/4} \right)z + 1}}{{{z^2} + 4}} \cr
& = \int {\frac{{z + 1}}{{z\left( {{z^2} + 4} \right)}}dz} = \int {\left( {\frac{1}{{4z}} + \frac{{ - \left( {1/4} \right)z + 1}}{{{z^2} + 4}}} \right)dx} \cr
& = \int {\frac{1}{{4z}}dz} - \frac{1}{4}\int {\frac{z}{{{z^2} + 4}}dz} + \int {\frac{1}{{{z^2} + 4}}dz} \cr
& {\text{integrating}} \cr
& = \frac{1}{4}\ln \left| z \right| - \frac{1}{4}\ln \left( {{z^2} + 4} \right) + {\tan ^{ - 1}}\left( z \right) + C \cr} $$
