Answer
$${\tan ^{ - 1}}\theta + \frac{2}{{{\theta ^2} + 1}} - \frac{1}{{4{{\left( {{\theta ^2} + 1} \right)}^2}}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{\theta ^4} - 4{\theta ^3} + 2{\theta ^2} - 3\theta + 1}}{{{{\left( {{\theta ^2} + 1} \right)}^3}}}} d\theta \cr
& {\text{Decompose the integrand }}\frac{{{\theta ^4} - 4{\theta ^3} + 2{\theta ^2} - 3\theta + 1}}{{{{\left( {{\theta ^2} + 1} \right)}^3}}}{\text{ into partial fractions}} \cr
& \frac{{{\theta ^4} - 4{\theta ^3} + 2{\theta ^2} - 3\theta + 1}}{{{{\left( {{\theta ^2} + 1} \right)}^3}}} = \frac{{A\theta + B}}{{{\theta ^2} + 1}} + \frac{{C\theta + D}}{{{{\left( {{\theta ^2} + 1} \right)}^2}}} + \frac{{E\theta + F}}{{{{\left( {{\theta ^2} + 1} \right)}^3}}} \cr
& {\text{Multiply by }}{\left( {{\theta ^2} + 1} \right)^3}{\text{ and simplify}} \cr
& {\theta ^4} - 4{\theta ^3} + 2{\theta ^2} - 3\theta + 1 = \left( {A\theta + B} \right){\left( {{\theta ^2} + 1} \right)^2} + \left( {C\theta + D} \right)\left( {{\theta ^2} + 1} \right) + E\theta + F \cr
& {\theta ^4} - 4{\theta ^3} + 2{\theta ^2} - 3\theta + 1 = \left( {A\theta + B} \right)\left( {{\theta ^4} + 2{\theta ^2} + 1} \right) + \left( {C\theta + D} \right)\left( {{\theta ^2} + 1} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + E\theta + F \cr
& {\theta ^4} - 4{\theta ^3} + 2{\theta ^2} - 3\theta + 1 = A{\theta ^5} + B{\theta ^4} + C{\theta ^3} + 2A{\theta ^3} + 2B{\theta ^2} + D{\theta ^2} + A\theta + C\theta \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + E\theta + B + D + F \cr
& {\text{Group terms}} \cr
& {\theta ^4} - 4{\theta ^3} + 2{\theta ^2} - 3\theta + 1 = A{\theta ^5} + B{\theta ^4} + \left( {C{\theta ^3} + 2A{\theta ^3}} \right) + \left( {2B{\theta ^2} + D{\theta ^2}} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( {A\theta + C\theta \, + E\theta } \right) + B + D + F \cr
& {\text{Equate coefficients}} \cr
& A = 0,\,\,\,\,\,B = 1,\,\,\,\,C + 2A = - 4,\,\,\,\,\,\,2B + D = 2,\,\,\,\,\,A + C + E = - 3,\,\,\,\,\,B + D + F = 1 \cr
& {\text{Solve the system of equations}} \cr
& A = 0,\,\,\,\,B = 1,\,\,\,C = - 4,\,\,\,D = 0,\,\,\,E = 1,\,\,\,F = 0 \cr
& {\text{Replace the coefficients}} \cr
& \frac{{{\theta ^4} - 4{\theta ^3} + 2{\theta ^2} - 3\theta + 1}}{{{{\left( {{\theta ^2} + 1} \right)}^3}}} = \frac{1}{{{\theta ^2} + 1}} + \frac{{ - 4\theta }}{{{{\left( {{\theta ^2} + 1} \right)}^2}}} + \frac{\theta }{{{{\left( {{\theta ^2} + 1} \right)}^3}}} \cr
& \int {\frac{{{\theta ^4} - 4{\theta ^3} + 2{\theta ^2} - 3\theta + 1}}{{{{\left( {{\theta ^2} + 1} \right)}^3}}}} d\theta = \int {\frac{1}{{{\theta ^2} + 1}}d\theta - 2\int {\frac{{2\theta }}{{{{\left( {{\theta ^2} + 1} \right)}^2}}}d\theta + \frac{1}{2}\int {\frac{{2\theta }}{{{{\left( {{\theta ^2} + 1} \right)}^3}}}} } } d\theta \cr
& {\text{Integrate}} \cr
& = {\tan ^{ - 1}}\theta + \frac{2}{{{\theta ^2} + 1}} + \frac{1}{2}\left( {\frac{{{{\left( {{\theta ^2} + 1} \right)}^{ - 2}}}}{{ - 2}}} \right) + C \cr
& = {\tan ^{ - 1}}\theta + \frac{2}{{{\theta ^2} + 1}} - \frac{1}{{4{{\left( {{\theta ^2} + 1} \right)}^2}}} + C \cr} $$
