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