Answer
$$\frac{1}{2}{\tan ^{ - 1}}x + \frac{1}{2}\left( {\frac{x}{{{x^2} + 1}}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{1 + 2{x^2} + {x^4}}}} \cr
& {\text{factor the perfect square}}{\text{, use }}{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} \cr
& = \int {\frac{{dx}}{{{{\left( {1 + {x^2}} \right)}^2}}}} \cr
& {\text{substitute }}x = \tan \theta ,{\text{ }}dx = {\sec ^2}\theta d\theta \cr
& = \int {\frac{{{{\sec }^2}\theta d\theta }}{{{{\left( {1 + {{\left( {\tan \theta } \right)}^2}} \right)}^2}}}} \cr
& = \int {\frac{{{{\sec }^2}\theta d\theta }}{{{{\left( {1 + {{\tan }^2}\theta } \right)}^2}}}} \cr
& {\text{identity }}1 + {\tan ^2}\theta = {\sec ^2}\theta \cr
& = \int {\frac{{{{\sec }^2}\theta d\theta }}{{{{\left( {{{\sec }^2}\theta } \right)}^2}}}} \cr
& = \int {\frac{{d\theta }}{{{{\sec }^2}\theta }}} \cr
& = \int {{{\cos }^2}\theta } d\theta \cr
& {\text{identity }}{\cos ^2}\theta = \frac{{1 + \cos 2\theta }}{2} \cr
& = \int {\frac{{1 + \cos 2\theta }}{2}} d\theta \cr
& {\text{find antiderivative}} \cr
& = \frac{\theta }{2} + \frac{1}{4}\sin 2\theta + C \cr
& {\text{double angle }}\sin 2\theta = 2\sin \theta \cos \theta \cr
& = \frac{\theta }{2} + \frac{1}{2}\sin \theta \cos \theta + C \cr
& {\text{write in terms of }}x,{\text{ }}x = \tan \theta \to \theta = {\tan ^{ - 1}}x,{\text{ }} \cr
& \sin \theta = \frac{x}{{\sqrt {{x^2} + 1} }},{\text{ and }}\cos \theta = \frac{1}{{\sqrt {{x^2} + 1} }} \cr
& = \frac{1}{2}{\tan ^{ - 1}}x + \frac{1}{2}\left( {\frac{x}{{\sqrt {{x^2} + 1} }}} \right)\left( {\frac{1}{{\sqrt {{x^2} + 1} }}} \right) + C \cr
& = \frac{1}{2}{\tan ^{ - 1}}x + \frac{1}{2}\left( {\frac{x}{{{x^2} + 1}}} \right) + C \cr} $$
