Answer
$\ln \left| {\frac{{x - 2}}{{x + 2}}} \right| + \frac{{x - 6}}{{{x^2} - 4}} + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{3{x^2} + 12x - 20}}{{{x^4} - 8{x^2} + 16}}} dx \cr
& {\text{Factoring the denominator}} \cr
& = \int {\frac{{3{x^2} + 12x - 20}}{{{{\left( {{x^2} - 4} \right)}^2}}}} dx \cr
& = \int {\frac{{3{x^2} + 12x - 20}}{{{{\left( {x - 2} \right)}^2}{{\left( {x + 2} \right)}^2}}}} dx \cr
& {\text{The partial fraction decomposition of the integrand has the}} \cr
& {\text{form}} \cr
& \frac{{3{x^2} + 12x - 20}}{{{{\left( {x - 2} \right)}^2}{{\left( {x + 2} \right)}^2}}} = \frac{A}{{x - 2}} + \frac{B}{{{{\left( {x - 2} \right)}^2}}} + \frac{C}{{x + 2}} + \frac{D}{{{{\left( {x + 2} \right)}^2}}} \cr
& {\text{Multiply both sides by the least common denominator}} \cr
& 3{x^2} + 12x - 20 = A\left( {x - 2} \right){\left( {x + 2} \right)^2} + B{\left( {x + 2} \right)^2} \cr
& + C\left( {x + 2} \right){\left( {x - 2} \right)^2} + D{\left( {x - 2} \right)^2} \cr
& {\text{Expanding and simplifying we obtain}} \cr
& 3{x^2} + 12x - 20 = A{x^3} + 2A{x^2} - 4Ax - 8A + B{x^2} + 4Bx + 4B \cr
& + C{x^3} - 2C{x^2} - 4Cx + 8C + D{x^2} - 4Dx + 4D \cr
& {\text{Collecting like terms}} \cr
& 3{x^2} + 12x - 20 = \left( {A + C} \right){x^3} + \left( {2A + B - 2C + D} \right){x^2} \cr
& + \left( { - 4A + 4B - 4C - 4D} \right)x + \left( { - 8A + 4B + 8C + 4D} \right) \cr
& {\text{We obtain the following system of equations}} \cr
& A + C = 0 \cr
& 2A + B - 2C + D = 3 \cr
& - 4A + 4B - 4C - 4D = 12 \cr
& - 8A + 4B + 8C + 4D = - 20 \cr
& {\text{Solving we get:}} \cr
& A = 1,{\text{ }}B = 1,{\text{ }}C = - 1,{\text{ }}D = - 2 \cr
& {\text{Therefore}}{\text{,}} \cr
& \frac{{3{x^2} + 12x - 20}}{{{{\left( {x - 2} \right)}^2}{{\left( {x + 2} \right)}^2}}} = \frac{A}{{x - 2}} + \frac{B}{{{{\left( {x - 2} \right)}^2}}} + \frac{C}{{x + 2}} + \frac{D}{{{{\left( {x + 2} \right)}^2}}} \cr
& \frac{{3{x^2} + 12x - 20}}{{{{\left( {x - 2} \right)}^2}{{\left( {x + 2} \right)}^2}}} = \frac{1}{{x - 2}} + \frac{1}{{{{\left( {x - 2} \right)}^2}}} - \frac{1}{{x + 2}} - \frac{2}{{{{\left( {x + 2} \right)}^2}}} \cr
& = \int {\left[ {\frac{1}{{x - 2}} + \frac{1}{{{{\left( {x - 2} \right)}^2}}} - \frac{1}{{x + 2}} - \frac{2}{{{{\left( {x + 2} \right)}^2}}}} \right]} dx \cr
& {\text{Integrating}} \cr
& = \ln \left| {x - 2} \right| - \frac{1}{{x - 2}} - \ln \left| {x + 2} \right| + \frac{2}{{x + 2}} + C \cr
& {\text{Simplifying}} \cr
& = \ln \left| {\frac{{x - 2}}{{x + 2}}} \right| + \frac{{x - 6}}{{{x^2} - 4}} + C \cr} $$