## Calculus: Early Transcendentals (2nd Edition)

$$2\ln \left| {\frac{{y - 1}}{{y + 1}}} \right| + \frac{5}{{y + 1}} - \frac{1}{{y - 1}} + C$$
\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}