Answer
$\frac{1}{4}\ln \left( {{x^4} + 4{x^2} + 3} \right) - \frac{1}{{2\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{x}{{\sqrt 3 }}} \right) - \frac{3}{2}{\tan ^{ - 1}}x + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^3} - 2{x^2} + 2x - 5}}{{{x^4} + 4{x^2} + 3}}} dx \cr
& {\text{Factoring the denominator}} \cr
& \int {\frac{{{x^3} - 2{x^2} + 2x - 5}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 1} \right)}}} dx \cr
& {\text{The partial fraction decomposition of the integrand has the}} \cr
& {\text{form}} \cr
& \frac{{{x^3} - 2{x^2} + 2x - 5}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 1} \right)}} = \frac{{Ax + B}}{{{x^2} + 3}} + \frac{{Cx + D}}{{{x^2} + 1}} \cr
& {\text{Multiply both sides by the least common denominator}} \cr
& {x^3} - 2{x^2} + 2x - 5 = \left( {Ax + B} \right)\left( {{x^2} + 1} \right) + \left( {Cx + D} \right)\left( {{x^2} + 3} \right) \cr
& {\text{Expanding and simplifying we obtain}} \cr
& {x^3} - 2{x^2} + 2x - 5 = A{x^3} + B{x^2} + Ax + B + C{x^3} + D{x^2} \cr
& + 3Cx + 3D \cr
& {\text{Collecting like terms}} \cr
& {x^3} - 2{x^2} + 2x - 5 = \left( {A + C} \right){x^3} + \left( {B + D} \right){x^2} + \left( {A + 3C} \right)x \cr
& + B + 3D \cr
& {\text{We obtain the following system of equations}} \cr
& A + C = 1 \cr
& B + D = - 2 \cr
& A + 3C = 2 \cr
& + B + 3D = - 5 \cr
& {\text{Solving we get:}} \cr
& A = \frac{1}{2},{\text{ }}B = - \frac{1}{2},{\text{ }}C = \frac{1}{2},{\text{ }}D = - \frac{3}{2} \cr
& {\text{Therefore}}{\text{,}} \cr
& \frac{{{x^3} - 2{x^2} + 2x - 5}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 1} \right)}} = \frac{{\left( {1/2} \right)x + \left( { - 1/2} \right)}}{{{x^2} + 3}} + \frac{{\left( {1/2} \right)x + \left( { - 3/2} \right)}}{{{x^2} + 1}} \cr
& \int {\frac{{{x^3} - 2{x^2} + 2x - 5}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 1} \right)}}} dx = \frac{1}{2}\int {\left( {\frac{x}{{{x^2} + 3}} - \frac{1}{{{x^2} + 3}}} \right)} dx \cr
& + \frac{1}{2}\int {\left( {\frac{x}{{{x^2} + 1}} - \frac{3}{{{x^2} + 1}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = \frac{1}{4}\ln \left( {{x^2} + 3} \right) - \frac{1}{{2\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{x}{{\sqrt 3 }}} \right) + \frac{1}{4}\ln \left( {{x^2} + 1} \right) - \frac{3}{2}{\tan ^{ - 1}}x + C \cr
& {\text{Simplifying}} \cr
& = \frac{1}{4}\ln \left( {{x^4} + 4{x^2} + 3} \right) - \frac{1}{{2\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{x}{{\sqrt 3 }}} \right) - \frac{3}{2}{\tan ^{ - 1}}x + C \cr} $$
