Answer
$$\ln \left| {x + 1} \right| - \frac{1}{2}\ln \left( {{x^2} + 2x + 2} \right) + \frac{1}{{2\left( {{x^2} + 2x + 2} \right)}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\left( {x + 1} \right){{\left( {{x^2} + 2x + 2} \right)}^2}}}} \cr
& {\text{Decompose }}\frac{1}{{\left( {x + 1} \right){{\left( {{x^2} + 2x + 2} \right)}^2}}}{\text{ into partial fractions}} \cr
& \frac{1}{{\left( {x + 1} \right){{\left( {{x^2} + 2x + 2} \right)}^2}}} = \frac{A}{{x + 1}} + \frac{{Bx + C}}{{{x^2} + 2x + 2}} + \frac{{Dx + E}}{{{{\left( {{x^2} + 2x + 2} \right)}^2}}} \cr
& {\text{Multiply both sides by }}\left( {x + 1} \right){\left( {{x^2} + 2x + 2} \right)^2} \cr
& 1 = A{\left( {{x^2} + 2x + 2} \right)^2} + \left( {Bx + C} \right)\left( {x + 1} \right)\left( {{x^2} + 2x + 2} \right) \cr
& \,\,\,\,\,\, + \left( {Dx + E} \right)\left( {x + 1} \right) \cr
& 1 = A\left( {{x^4} + 4{x^3} + 8{x^2} + 8x + 4} \right) + \left( {Bx + C} \right)\left( {{x^3} + 3{x^2} + 4x + 2} \right) \cr
& \,\,\,\,\, + D{x^2} + Dx + Ex + E \cr
& 1 = A{x^4} + 4A{x^3} + 8A{x^2} + 8Ax + 4A + B{x^4} + C{x^3} + 3B{x^3} \cr
& \,\,\,\,\,\, + 3C{x^2} + 4B{x^2} + 4Cx + 2Bx + 2C\, + D{x^2} + Dx + Ex + E \cr
& {\text{Group like terms}} \cr
& 1 = \left( {A + B} \right){x^4} + \left( {4A + C + 3B} \right){x^3} + \left( {8A + 3C + 4B + D} \right){x^2} \cr
& \,\,\,\,\, + \left( {8A + 4C + 2B + D + E} \right)x + \left( {4A + 2C + E\,} \right) \cr
& {\text{We obtain the system of linear equations}} \cr
& A + B = 0 \cr
& 4A + C + 3B = 0 \cr
& 8A + 3C + 4B + D = 0 \cr
& 8A + 4C + 2B + D + E = 0 \cr
& 4A + 2C + E\, = 1 \cr
& {\text{Solving the system using a CAS we obtain}} \cr
& A = 1,\,\,\,B = - 1,\,\,\,\,C = - 1,\,\,\,\,D = - 1,\,\,\,\,E = - 1 \cr
& \frac{1}{{\left( {x + 1} \right){{\left( {{x^2} + 2x + 2} \right)}^2}}} = \frac{1}{{x + 1}} - \frac{{x + 1}}{{{x^2} + 2x + 2}} - \frac{{x + 1}}{{{{\left( {{x^2} + 2x + 2} \right)}^2}}} \cr
& \cr
& \int {\frac{{dx}}{{\left( {x + 1} \right){{\left( {{x^2} + 2x + 2} \right)}^2}}}} = \int {\frac{1}{{x + 1}}} dx - \int {\frac{{x + 1}}{{{x^2} + 2x + 2}}dx} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \int {\frac{{x + 1}}{{{{\left( {{x^2} + 2x + 2} \right)}^2}}}dx} \cr
& {\text{Integrate}} \cr
& = \ln \left| {x + 1} \right| - \frac{1}{2}\ln \left( {{x^2} + 2x + 2} \right) + \frac{1}{{2\left( {{x^2} + 2x + 2} \right)}} + C \cr} $$
