Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.5 Exercises - Page 549: 18

$$\ln \left| {x - 2} \right| - \frac{1}{2}\ln \left( {{x^2} + 2x + 4} \right) + \sqrt 3 \arctan \left( {\frac{{x + 1}}{{\sqrt 3 }}} \right) + C$$

$$\eqalign{ & \int {\frac{{6x}}{{{x^3} - 8}}} dx \cr & {\text{Factor the denominator}} \cr & \int {\frac{{6x}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}} dx \cr & {\text{Decompose the integrand into partial fractions}} \cr & \frac{{6x}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}} = \frac{A}{{x - 2}} + \frac{{Bx + C}}{{{x^2} + 2x + 4}} \cr & 6x = A\left( {{x^2} + 2x + 4} \right) + \left( {Bx + C} \right)\left( {x - 2} \right) \cr & 6x = A{x^2} + 2Ax + 4A + B{x^2} - 2Bx + Cx - 2C \cr & 6x = A{x^2} + B{x^2} + 2Ax - 2Bx + Cx + 4A - 2C \cr & {\text{Equating coefficients}} \cr & A + B = 0 \cr & 2A - 2B + C = 6 \cr & 4A - 2C = 0 \cr & {\text{Solving the system of equations using a calculator}} \cr & A = 1,{\text{ }}B = - 1,{\text{ }}C = 2 \cr & \frac{{6x}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}} = \frac{A}{{x - 2}} + \frac{{Bx + C}}{{{x^2} + 2x + 4}} \cr & \frac{{6x}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}} = \frac{1}{{x - 2}} + \frac{{ - x + 2}}{{{x^2} + 2x + 4}} \cr & \int {\frac{{6x}}{{{x^3} - 8}}} dx = \int {\frac{1}{{x - 2}}} dx - \int {\frac{{x - 2}}{{{x^2} + 2x + 4}}} dx \cr & {\text{Completing the square}} \cr & = \int {\frac{1}{{x - 2}}} dx - \int {\frac{{x - 2}}{{\left( {{x^2} + 2x + 1} \right) + 3}}} dx \cr & = \int {\frac{1}{{x - 2}}} dx - \int {\frac{{x - 2}}{{{{\left( {x + 1} \right)}^2} + 3}}} dx \cr & = \ln \left| {x - 2} \right| - \int {\frac{{x - 2}}{{{{\left( {x + 1} \right)}^2} + 3}}} dx{\text{ }}\left( {\bf{1}} \right) \cr & {\text{Integrating }}\int {\frac{{x - 2}}{{{{\left( {x + 1} \right)}^2} + 3}}} dx \cr & {\text{Let }}u = x + 1,{\text{ }}x = u - 1,{\text{ }}dx = du \cr & \int {\frac{{x - 2}}{{{{\left( {x + 1} \right)}^2} + 3}}} dx = \int {\frac{{u - 1 - 2}}{{{u^2} + 3}}} du \cr & = \int {\frac{{u - 3}}{{{u^2} + 3}}} du \cr & = \int {\frac{u}{{{u^2} + 3}}} du - \int {\frac{3}{{{u^2} + 3}}} \cr & = \frac{1}{2}\ln \left( {{u^2} + 3} \right) - \frac{3}{{\sqrt 3 }}\arctan \left( {\frac{u}{{\sqrt 3 }}} \right) + C \cr & = \frac{1}{2}\ln \left( {{x^2} + 2x + 4} \right) - \sqrt 3 \arctan \left( {\frac{{x + 1}}{{\sqrt 3 }}} \right) + C \cr & {\text{Substitute the previous result into }}\left( {\bf{1}} \right) \cr & = \ln \left| {x - 2} \right| - \frac{1}{2}\ln \left( {{x^2} + 2x + 4} \right) + \sqrt 3 \arctan \left( {\frac{{x + 1}}{{\sqrt 3 }}} \right) + C \cr} $$