Answer
$ - \frac{3}{2} - \ln 2$
Work Step by Step
$$\eqalign{
& \int_{ - 2}^{ - 1} {\frac{{{z^2} + 1}}{z}} dz \cr
& {\text{Distribute the numerator in the integrand}} \cr
& = \int_{ - 2}^{ - 1} {\left( {\frac{{{z^2}}}{z} + \frac{1}{z}} \right)} dz \cr
& = \int_{ - 2}^{ - 1} {\left( {z + \frac{1}{z}} \right)} dz \cr
& {\text{Integrate}}{\text{, use }}\int {{x^n}dx} = \frac{{{x^{n + 1}}}}{{n + 1}} + C{\text{ and }}\int {\frac{1}{x}} dx = \ln \left| x \right| + C \cr
& = \left[ {\frac{{{z^2}}}{2} + \ln \left| z \right|} \right]_{ - 2}^{ - 1} \cr
& {\text{Evaluate the limits and simplify}} \cr
& = \left[ {\frac{{{{\left( { - 1} \right)}^2}}}{2} + \ln \left| { - 1} \right|} \right] - \left[ {\frac{{{{\left( { - 2} \right)}^2}}}{2} + \ln \left| { - 2} \right|} \right] \cr
& = \left( {\frac{1}{2} + 0} \right) - \left( {2 + \ln 2} \right) \cr
& = \frac{1}{2} - 2 - \ln 2 \cr
& = - \frac{3}{2} - \ln 2 \cr} $$
