## Calculus: Early Transcendentals (2nd Edition)

$$2\ln \left| {\frac{{x - 2}}{{x + 2}}} \right| + \ln \left| {\frac{{x + 1}}{{x - 1}}} \right| + C$$
\eqalign{ & \int {\frac{{6{x^2}}}{{{x^4} - 5{x^2} + 4}}dx} \cr & {\text{integrand }} \cr & = \frac{{6{x^2}}}{{{x^4} - 5{x^2} + 4}} \cr & {\text{factor}} \cr & = \frac{{6{x^2}}}{{\left( {{x^2} - 4} \right)\left( {{x^2} - 1} \right)}} \cr & = \frac{{6{x^2}}}{{\left( {x + 2} \right)\left( {x - 2} \right)\left( {x + 1} \right)\left( {x - 1} \right)}} \cr & {\text{partial fractions}} \cr & \frac{{6{x^2}}}{{\left( {x + 2} \right)\left( {x - 2} \right)\left( {x + 1} \right)\left( {x - 1} \right)}} = \frac{A}{{x + 2}} + \frac{B}{{x - 2}} + \frac{C}{{x + 1}} + \frac{D}{{x - 1}} \cr & 6{x^2} = A\left( {x - 2} \right)\left( {x + 1} \right)\left( {x - 1} \right) + B\left( {x + 2} \right)\left( {x + 1} \right)\left( {x - 1} \right) + C\left( {x + 2} \right)\left( {x - 2} \right)\left( {x - 1} \right) \cr & + D\left( {x + 2} \right)\left( {x - 2} \right)\left( {x + 1} \right) \cr & {\text{Letting }}x = - 2 \cr & 6{\left( { - 2} \right)^2} = A\left( { - 2 - 2} \right)\left( { - 2 + 1} \right)\left( { - 2 - 1} \right) + B\left( 0 \right) + C\left( 0 \right) + D\left( 0 \right) \cr & 24 = A\left( { - 12} \right) \cr & A = - 2 \cr & {\text{Letting }}x = 2 \cr & 6{\left( 2 \right)^2} = A\left( 0 \right) + B\left( {2 + 2} \right)\left( {2 + 1} \right)\left( {2 - 1} \right) + C\left( 0 \right) + D\left( 0 \right) \cr & 24 = B\left( {12} \right) \cr & B = 2 \cr & {\text{Letting }}x = - 1 \cr & 6{\left( { - 1} \right)^2} = A\left( 0 \right) + B\left( 0 \right) + C\left( { - 1 + 2} \right)\left( { - 1 - 2} \right)\left( { - 1 - 1} \right) + D\left( 0 \right) \cr & 6 = C\left( 6 \right) \cr & C = 1 \cr & {\text{Letting }}x = 1 \cr & 6{\left( 1 \right)^2} = A\left( 0 \right) + B\left( 0 \right) + C\left( 0 \right) + D\left( {1 + 2} \right)\left( {1 - 2} \right)\left( {1 + 1} \right) \cr & 6 = D\left( { - 6} \right) \cr & D = - 1 \cr & {\text{substituting constants}} \cr & \frac{A}{{x + 2}} + \frac{B}{{x - 2}} + \frac{C}{{x + 1}} + \frac{D}{{x - 1}} = \frac{{ - 2}}{{x + 2}} + \frac{2}{{x - 2}} + \frac{1}{{x + 1}} + \frac{{ - 1}}{{x - 1}} \cr & \int {\frac{{6{x^2}}}{{{x^4} - 5{x^2} + 4}}dx} = \int {\left( {\frac{{ - 2}}{{x + 2}} + \frac{2}{{x - 2}} + \frac{1}{{x + 1}} + \frac{{ - 1}}{{x - 1}}} \right)dx} \cr & {\text{integrating}} \cr & = - 2\ln \left| {x + 2} \right| + 2\ln \left| {x - 2} \right| + \ln \left| {x + 1} \right| - \ln \left| {x - 1} \right| + C \cr & {\text{simplify}} \cr & = 2\ln \left| {\frac{{x - 2}}{{x + 2}}} \right| + \ln \left| {\frac{{x + 1}}{{x - 1}}} \right| + C \cr}