Answer
$$\frac{A}{{x - 2}} + \frac{{Bx + C}}{{{x^2} + 1}} + \frac{{Dx + E}}{{{{\left( {{x^2} + 1} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& \frac{{1 - 3{x^4}}}{{\left( {x - 2} \right){{\left( {{x^2} + 1} \right)}^2}}} \cr
& x - 2{\text{ is a linear factor, then its decomposition is }}\frac{A}{{x - 2}} \cr
& {\left( {{x^2} + 1} \right)^2}{\text{ is a linear repeated quadratic factor, then}} \cr
& {\text{the numerator for each term it is a linear factor }}Bx + C{\text{ }} \cr
& {\text{and }}Dx + E \cr
& {\text{The partial decomposition is:}} \cr
& \frac{{1 - 3{x^4}}}{{\left( {x - 2} \right){{\left( {{x^2} + 1} \right)}^2}}} = \frac{A}{{x - 2}} + \frac{{Bx + C}}{{{x^2} + 1}} + \frac{{Dx + E}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr} $$
