Answer
$$\ln \left( {{x^2} + 2x + 2} \right) - 7\arctan \left( {x + 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{2x - 5}}{{{x^2} + 2x + 2}}} dx \cr
& {\text{Completing the square}} \cr
& \int {\frac{{2x - 5}}{{{x^2} + 2x + 2}}} dx = \int {\frac{{2x - 5}}{{\left( {{x^2} + 2x + 1} \right) + 1}}} dx \cr
& = \int {\frac{{2x - 5}}{{{{\left( {x + 1} \right)}^2} + 1}}} dx \cr
& {\text{Let }}u = x + 1,{\text{ }}x = u - 1,{\text{ }}dx = du,{\text{ }} \cr
& \int {\frac{{2x - 5}}{{{{\left( {x + 1} \right)}^2} + 1}}} dx = \int {\frac{{2\left( {u - 1} \right) - 5}}{{{u^2} + 1}}} du \cr
& = \int {\frac{{2u - 7}}{{{u^2} + 1}}} du \cr
& {\text{Distribute}} \cr
& = \int {\frac{{2u}}{{{u^2} + 1}}} du - \int {\frac{7}{{{u^2} + 1}}} du \cr
& {\text{Integrate using basic integration rules}} \cr
& = \ln \left( {{u^2} + 1} \right) - 7\arctan u + C \cr
& {\text{Write in terms of }}x \cr
& = \ln \left[ {{{\left( {x + 1} \right)}^2} + 1} \right] - 7\arctan \left( {x + 1} \right) + C \cr
& = \ln \left( {{x^2} + 2x + 2} \right) - 7\arctan \left( {x + 1} \right) + C \cr} $$