Answer
$$\ln \left| {x + 1} \right| + \sqrt 2 {\tan ^{ - 1}}\left( {\frac{{x - 1}}{{\sqrt 2 }}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2} + 5}}{{{x^3} - {x^2} + x + 3}}} dx \cr
& {\text{Factor the denominator}} \cr
& = \int {\frac{{{x^2} + 5}}{{\left( {x + 1} \right)\left( {{x^2} - 2x + 3} \right)}}} dx \cr
& {\text{Decompose the integrand into partial fractions}} \cr
& \frac{{{x^2} + 5}}{{\left( {x + 1} \right)\left( {{x^2} - 2x + 3} \right)}} = \frac{A}{{x + 1}} + \frac{{Bx + C}}{{{x^2} - 2x + 3}} \cr
& {x^2} + 5 = A\left( {{x^2} - 2x + 3} \right) + \left( {Bx + C} \right)\left( {x + 1} \right) \cr
& {x^2} + 5 = A{x^2} - 2Ax + 3A + B{x^2} + Bx + Cx + C \cr
& {x^2} + 5 = \left( {A{x^2} + B{x^2}} \right) + \left( { - 2Ax + Bx + Cx} \right) + \left( {3A + C} \right) \cr
& {\text{Setting up a system of equations}} \cr
& A + B = 1 \cr
& - 2A + B + C = 0 \cr
& 3A + C = 5 \cr
& {\text{Solving the system of equations by using a CAS or calculator}} \cr
& A = 1,{\text{ }}B = 0,{\text{ }}C = 2 \cr
& \frac{{{x^2} + 5}}{{\left( {x + 1} \right)\left( {{x^2} - 2x + 3} \right)}} = \frac{1}{{x + 1}} + \frac{2}{{{x^2} - 2x + 3}} \cr
& {\text{Completing the square}} \cr
& \frac{{{x^2} + 5}}{{\left( {x + 1} \right)\left( {{x^2} - 2x + 3} \right)}} = \frac{1}{{x + 1}} + \frac{2}{{{x^2} - 2x + 1 + 2}} \cr
& \frac{{{x^2} + 5}}{{\left( {x + 1} \right)\left( {{x^2} - 2x + 3} \right)}} = \frac{1}{{x + 1}} + \frac{2}{{{{\left( {x - 1} \right)}^2} + 2}} \cr
& \int {\frac{{{x^2} + 5}}{{\left( {x + 1} \right)\left( {{x^2} - 2x + 3} \right)}}} dx = \int {\frac{1}{{x + 1}}} dx + \int {\frac{2}{{{{\left( {x - 1} \right)}^2} + 2}}dx} \cr
& {\text{Integrating}} \cr
& = \ln \left| {x + 1} \right| + 2\left[ {\frac{1}{{\sqrt 2 }}{{\tan }^{ - 1}}\left( {\frac{{x - 1}}{{\sqrt 2 }}} \right)} \right] + C \cr
& = \ln \left| {x + 1} \right| + \sqrt 2 {\tan ^{ - 1}}\left( {\frac{{x - 1}}{{\sqrt 2 }}} \right) + C \cr} $$
