Answer
$$\frac{2}{3}\ln \left| {1 + x} \right| + \frac{1}{6}\ln \left( {{x^2} - x + 1} \right) + \frac{{\sqrt 3 }}{3}{\tan ^{ - 1}}\left( {\frac{{2x - 1}}{{\sqrt 3 }}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{1 + {x^2}}}{{1 + {x^3}}}} dx \cr
& {\text{Decomposing the integrand into partial fractions}} \cr
& \frac{{1 + {x^2}}}{{1 + {x^3}}} = \frac{{1 + {x^2}}}{{\left( {1 + x} \right)\left( {1 - x + {x^2}} \right)}} = \frac{A}{{1 + x}} + \frac{{Bx + C}}{{1 - x + {x^2}}} \cr
& \,\,\,{\text{Multiply by }}\left( {1 + x} \right)\left( {1 - x + {x^2}} \right) \cr
& \,\,\,1 + {x^2} = A\left( {1 - x + {x^2}} \right) + \left( {Bx + C} \right)\left( {1 + x} \right) \cr
& \,\,\,1 + {x^2} = A - Ax + A{x^2} + Bx + B{x^2} + C + Cx \cr
& \,\,\,1 + {x^2} = \left( {A{x^2} + B{x^2}} \right) + \left( { - Ax + Bx + Cx} \right) + A + C \cr
& \,\,{\text{Equating coefficients }}A + B = 1,\,\,\,\, - A + B + C = 0,\,{\text{ }}A + C = 1 \cr
& \,\,{\text{Solving the system of equations with a calculator, we obtain}} \cr
& \,\,\,A = \frac{2}{3},\,\,B = \frac{1}{3},\,\,C = \frac{1}{3},\,\, \cr
& {\text{then}}{\text{,}} \cr
& \,\,\,\,\frac{{1 + {x^2}}}{{\left( {1 + x} \right)\left( {1 - x + {x^2}} \right)}} = \frac{A}{{1 + x}} + \frac{{Bx + C}}{{1 - x + {x^2}}} = \frac{2}{{3\left( {1 + x} \right)}} + \frac{{x + 1}}{{3\left( {1 - x + {x^2}} \right)}} \cr
& \cr
& \int {\frac{{1 + {x^2}}}{{1 + {x^3}}}} dx = \int {\frac{2}{{3\left( {1 + x} \right)}}} dx + \int {\frac{{x + 1}}{{3\left( {1 - x + {x^2}} \right)}}} dx \cr
& = \frac{2}{3}\int {\frac{1}{{1 + x}}} dx + \frac{1}{3}\int {\frac{{x + 1}}{{{x^2} - x + 1}}} dx \cr
& \cr
& {\text{Integrating }}\int {\frac{{x + 1}}{{{x^2} - x + 1}}} dx{\text{ by completing the square}} \cr
& \int {\frac{{x + 1}}{{{x^2} - x + 1}}} dx = \int {\frac{{x + 1}}{{{x^2} - x + \frac{1}{4} + \frac{3}{4}}}} dx = \int {\frac{{x + 1}}{{{{\left( {x - 1/2} \right)}^2} + {{\left( {\sqrt 3 /2} \right)}^2}}}} dx \cr
& u = x - 1/2,\,\,\,du = dx \cr
& = \int {\frac{{u + 3/2}}{{{u^2} + {{\left( {\sqrt 3 /2} \right)}^2}}}} du = \int {\frac{u}{{{u^2} + {{\left( {\sqrt 3 /2} \right)}^2}}}} du + \int {\frac{{3/2}}{{{u^2} + {{\left( {\sqrt 3 /2} \right)}^2}}}} du \cr
& = \frac{1}{2}\ln \left( {{u^2} + {{\left( {\sqrt 3 /2} \right)}^2}} \right) + \frac{3}{2}\left( {\frac{2}{{\sqrt 3 }}} \right){\tan ^{ - 1}}\left( {\frac{{2u}}{{\sqrt 3 }}} \right) + C \cr
& {\text{replace }}u = x - 1/2 \cr
& = \frac{1}{2}\ln \left( {{x^2} - x + 1} \right) + \frac{3}{2}\left( {\frac{2}{{\sqrt 3 }}} \right){\tan ^{ - 1}}\left( {\frac{{2\left( {x - 1/2} \right)}}{{\sqrt 3 }}} \right) + C \cr
& = \frac{1}{2}\ln \left( {{x^2} - x + 1} \right) + \sqrt 3 {\tan ^{ - 1}}\left( {\frac{{2x - 1}}{{\sqrt 3 }}} \right) + C \cr
& {\text{Then}}{\text{,}} \cr
& \frac{2}{3}\int {\frac{1}{{1 + x}}} dx + \frac{1}{3}\int {\frac{{x + 1}}{{{x^2} - x + 1}}} dx \cr
& \to = \frac{2}{3}\ln \left| {1 + x} \right| + \frac{1}{3}\left( {\frac{1}{2}\ln \left( {{x^2} - x + 1} \right) + \sqrt 3 {{\tan }^{ - 1}}\left( {\frac{{2x - 1}}{{\sqrt 3 }}} \right)} \right) + C \cr
& = \frac{2}{3}\ln \left| {1 + x} \right| + \frac{1}{6}\ln \left( {{x^2} - x + 1} \right) + \frac{{\sqrt 3 }}{3}{\tan ^{ - 1}}\left( {\frac{{2x - 1}}{{\sqrt 3 }}} \right) + C \cr} $$
