Answer
$$ - \frac{1}{{{{\left( {1 + x} \right)}^2}}} + \ln \left| {1 + x} \right| + \frac{2}{{1 + x}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{1 + {x^2}}}{{{{\left( {1 + x} \right)}^3}}}} dx \cr
& {\text{Integrate by using the substitution method}} \cr
& \,\,\,{\text{Let }}u = 1 + x,\,\,\,\,\,\,x = u - 1,\,\,\,\,dx = du \cr
& \cr
& {\text{Write the integrand in terms of }}u \cr
& \int {\frac{{1 + {x^2}}}{{{{\left( {1 + x} \right)}^3}}}} dx = \int {\frac{{1 + {{\left( {u - 1} \right)}^2}}}{{{u^3}}}} du \cr
& {\text{expand and distribute}} \cr
& = \int {\frac{{1 + {u^2} - 2u + 1}}{{{u^3}}}} du \cr
& = \int {\frac{{2 + {u^2} - 2u}}{{{u^3}}}} du \cr
& = \int {\left( {\frac{2}{{{u^3}}} + \frac{{{u^2}}}{{{u^3}}} - \frac{{2u}}{{{u^3}}}} \right)} du \cr
& = \int {\left( {2{u^{ - 3}} + \frac{1}{u} - 2{u^{ - 2}}} \right)} du \cr
& \cr
& {\text{Integrating}} \cr
& = 2\left( {\frac{{{u^{ - 2}}}}{{ - 2}}} \right) + \ln \left| u \right| - 2\left( {\frac{{{u^{ - 1}}}}{{ - 1}}} \right) + C \cr
& = - \frac{1}{{{u^2}}} + \ln \left| u \right| + \frac{2}{u} + C \cr
& \cr
& {\text{Write in terms of }}x,{\text{ substitute }}1 + x{\text{ for }}u \cr
& = - \frac{1}{{{{\left( {1 + x} \right)}^2}}} + \ln \left| {1 + x} \right| + \frac{2}{{1 + x}} + C \cr} $$
