Answer
$${z^2} - z + 4\ln \left| {z + 3} \right| + 3\ln \left| {z - 2} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{2{z^3} + {z^2} - 6z + 7}}{{{z^2} + z - 6}}dz} \cr
& {\text{use long division}} \cr
& = \int {\left( {2z - 1 + \frac{{7z + 1}}{{{z^2} + z - 6}}} \right)} dz \cr
& {\text{partial fractions}} \cr
& \frac{{7z + 1}}{{{z^2} + z - 6}} = \frac{{7z + 1}}{{\left( {z + 3} \right)\left( {z - 2} \right)}} \cr
& \frac{{7z + 1}}{{\left( {z + 3} \right)\left( {z - 2} \right)}} = \frac{A}{{z + 3}} + \frac{B}{{z - 2}} \cr
& 7z + 1 = A\left( {z - 2} \right) + B\left( {z + 3} \right) \cr
& z = - 3 \to 4 = A \cr
& z = 2 \to 3 = B \cr
& \frac{{7z + 1}}{{\left( {z + 3} \right)\left( {z - 2} \right)}} = \frac{4}{{z + 3}} + \frac{3}{{z - 2}} \cr
& \int {\left( {2z - 1 + \frac{{7z + 1}}{{{z^2} + z - 6}}} \right)} dz = \int {\left( {2z - 1 + \frac{4}{{z + 3}} + \frac{3}{{z - 2}}} \right)} dx \cr
& {\text{find the antiderivative}} \cr
& = {z^2} - z + 4\ln \left| {z + 3} \right| + 3\ln \left| {z - 2} \right| + C \cr} $$
