Answer
$$2\ln \left| {\frac{{y - 1}}{{y + 1}}} \right| + \frac{5}{{y + 1}} - \frac{1}{{y - 1}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{12y - 8}}{{{y^4} - 2{y^2} + 1}}dx} \cr
& {\text{integrand }} \cr
& = \frac{{12y - 8}}{{{y^4} - 2{y^2} + 1}} \cr
& {\text{factor}} \cr
& = \frac{{12y - 8}}{{{{\left( {{y^2} - 1} \right)}^2}}} = \frac{{12y - 8}}{{{{\left( {y + 1} \right)}^2}{{\left( {y - 1} \right)}^2}}} \cr
& {\text{partial fractions}} \cr
& \frac{{12y - 8}}{{{{\left( {y + 1} \right)}^2}{{\left( {y - 1} \right)}^2}}} = \frac{A}{{y + 1}} + \frac{B}{{{{\left( {y + 1} \right)}^2}}} + \frac{C}{{y - 1}} + \frac{D}{{{{\left( {y - 1} \right)}^2}}} \cr
& 12y - 8 = A\left( {y + 1} \right){\left( {y - 1} \right)^2} + B{\left( {y - 1} \right)^2} + C\left( {y - 1} \right){\left( {y + 1} \right)^2} + D{\left( {y + 1} \right)^2} \cr
& {\text{multiplying}} \cr
& 12y - 8 = A\left( {{y^3} - {y^2} - y + 1} \right) + B\left( {{y^2} - 2y + 1} \right) + C\left( {{y^3} + {y^2} - y - 1} \right) + D\left( {{y^2} + 2y + 1} \right) \cr
& 12y - 8 = A{y^3} - A{y^2} - Ay + A + B{y^2} - 2By + B + C{y^3} + C{y^2} - Cy - C \cr
& + D{y^2} + 2Dy + D \cr
& 12y - 8 = \left( {A{y^3} + C{y^3}} \right) + \left( { - A{y^2} + B{y^2} + C{y^2} + D{y^2}} \right) + \left( { - Ay - 2By - Cy + 2Dy} \right) + \cr
& A + B - C + D \cr
& {\text{by equating the coefficients}} \cr
& {x^3}:{\text{ }}A + C = 0 \cr
& {x^2}:{\text{ }} - A + B + C + D = 0 \cr
& x:{\text{ }} - A - 2B - C + 2D = 12 \cr
& {x^0}:{\text{ }}A + B - C + D = - 8 \cr
& {\text{Solving these equations}} \cr
& A = - 2 \cr
& B = - 5 \cr
& C = 2 \cr
& D = 1 \cr
& {\text{substituting constants}} \cr
& \frac{A}{{y + 1}} + \frac{B}{{{{\left( {y + 1} \right)}^2}}} + \frac{C}{{y - 1}} + \frac{D}{{{{\left( {y - 1} \right)}^2}}} = \frac{{ - 2}}{{y + 1}} + \frac{{ - 5}}{{{{\left( {y + 1} \right)}^2}}} + \frac{2}{{y - 1}} + \frac{1}{{{{\left( {y - 1} \right)}^2}}} \cr
& \int {\frac{{12y - 8}}{{{y^4} - 2{y^2} + 1}}dx} = \int {\left( {\frac{{ - 2}}{{y + 1}} + \frac{{ - 5}}{{{{\left( {y + 1} \right)}^2}}} + \frac{2}{{y - 1}} + \frac{1}{{{{\left( {y - 1} \right)}^2}}}} \right)dx} \cr
& {\text{integrating}} \cr
& = - 2\ln \left| {y + 1} \right| + \frac{5}{{y + 1}} + 2\ln \left| {y - 1} \right| - \frac{1}{{y - 1}} + C \cr
& {\text{simplify}} \cr
& = 2\ln \left| {\frac{{y - 1}}{{y + 1}}} \right| + \frac{5}{{y + 1}} - \frac{1}{{y - 1}} + C \cr} $$
