Answer
$$\frac{1}{4}\ln \left( {{x^4} + 2{x^2} + 1} \right) + \frac{1}{2}{\tan ^{ - 1}}x + \frac{x}{{2\left( {{x^2} + 1} \right)}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^3} + x + 1}}{{{x^4} + 2{x^2} + 1}}} dx \cr
& {\text{Distribute the numerator}} \cr
& \int {\frac{{{x^3} + x + 1}}{{{x^4} + 2{x^2} + 1}}} dx = \int {\frac{{{x^3} + x}}{{{x^4} + 2{x^2} + 1}}} dx + \int {\frac{1}{{{x^4} + 2{x^2} + 1}}} dx \cr
& = \frac{1}{4}\int {\frac{{4{x^3} + 4x}}{{{x^4} + 2{x^2} + 1}}} dx + \int {\frac{1}{{{x^4} + 2{x^2} + 1}}} dx \cr
& = \frac{1}{4}\ln \left( {{x^4} + 2{x^2} + 1} \right) + \int {\frac{1}{{{x^4} + 2{x^2} + 1}}} dx{\text{ }}\left( {\bf{1}} \right) \cr
& \cr
& {\text{*Integrating }}\int {\frac{1}{{{x^4} + 2{x^2} + 1}}} dx \cr
& {\text{Factor the denominator}} \cr
& \int {\frac{1}{{{x^4} + 2{x^2} + 1}}} dx = \int {\frac{1}{{{{\left( {{x^2} + 1} \right)}^2}}}} dx \cr
& {\text{Using the substitution }}\tan \theta = x,{\text{ }}dx = {\sec ^2}\theta d\theta \cr
& \int {\frac{1}{{{{\left( {{x^2} + 1} \right)}^2}}}} dx = \int {\frac{{{{\sec }^2}\theta }}{{{{\left( {{{\tan }^2}\theta + 1} \right)}^2}}}} d\theta \cr
& = \int {\frac{{{{\sec }^2}\theta }}{{{{\left( {{{\sec }^2}\theta } \right)}^2}}}} d\theta = \int {\frac{1}{{{{\sec }^2}\theta }}} d\theta = \int {{{\cos }^2}\theta } d\theta \cr
& = \int {\left( {\frac{1}{2} + \frac{1}{2}\cos 2\theta } \right)} d\theta \cr
& {\text{Integrating}} \cr
& = \frac{1}{2}\theta + \frac{1}{4}\sin 2\theta + C \cr
& = \frac{1}{2}\theta + \frac{1}{4}\left( {2\sin \theta \cos \theta } \right) + C \cr
& = \frac{1}{2}\theta + \frac{1}{2}\sin \theta \cos \theta + C \cr
& {\text{We know that }}\tan \theta = \frac{x}{1},{\text{ }} \cr
& {\text{sin}}\theta = \frac{x}{{\sqrt {{x^2} + 1} }}{\text{ and cos}}\theta = \frac{1}{{\sqrt {{x^2} + 1} }} \cr
& = \frac{1}{2}{\tan ^{ - 1}}x + \frac{1}{2}\left( {\frac{x}{{\sqrt {{x^2} + 1} }}} \right)\left( {\frac{1}{{\sqrt {{x^2} + 1} }}} \right) + C \cr
& = \frac{1}{2}{\tan ^{ - 1}}x + \frac{x}{{2\left( {{x^2} + 1} \right)}} + C \cr
& {\text{Substituting into }}\left( {\bf{1}} \right) \cr
& = \frac{1}{4}\ln \left( {{x^4} + 2{x^2} + 1} \right) + \frac{1}{2}{\tan ^{ - 1}}x + \frac{x}{{2\left( {{x^2} + 1} \right)}} + C \cr} $$
