## Calculus: Early Transcendentals (2nd Edition)

$$2\ln \left| {x + 1} \right| + {\tan ^{ - 1}}\left( {x + 1} \right) + C$$
\eqalign{ & \int {\frac{{2{x^2} + 5x + 5}}{{\left( {x + 1} \right)\left( {{x^2} + 2x + 2} \right)}}dx} \cr & {\text{partial fractions}} \cr & \frac{{2{x^2} + 5x + 5}}{{\left( {x + 1} \right)\left( {{x^2} + 2x + 2} \right)}} = \frac{A}{{x + 1}} + \frac{{Bx + C}}{{{x^2} + 2x + 2}} \cr & {\text{multiplying}} \cr & 2{x^2} + 5x + 5 = A\left( {{x^2} + 2x + 2} \right) + \left( {Bx + C} \right)\left( {x + 1} \right) \cr & 2{x^2} + 5x + 5 = A{x^2} + 2Ax + 2A + B{x^2} + Bx + Cx + C \cr & 2{x^2} + 5x + 5 = \left( {A{x^2} + B{x^2}} \right) + \left( {2Ax + Bx + Cx} \right) + \left( {2A + C} \right) \cr & {\text{by equating the coefficients}} \cr & {x^2}:{\text{ }}A + B = 2 \cr & x:{\text{ 2}}A + B + C = 5 \cr & {x^0}:{\text{ }}2A + C = 5 \cr & {\text{Solving these equations}} \cr & A = 2 \cr & B = 0 \cr & C = 1 \cr & {\text{substituting constants}} \cr & \frac{A}{{x + 1}} + \frac{{Bx + C}}{{{x^2} + 2x + 2}} = \frac{2}{{x + 1}} + \frac{1}{{{x^2} + 2x + 2}} \cr & = \int {\frac{{2{x^2} + 5x + 5}}{{\left( {x + 1} \right)\left( {{x^2} + 2x + 2} \right)}}dx} = \int {\left( {\frac{2}{{x + 1}} + \frac{1}{{{x^2} + 2x + 2}}} \right)dx} \cr & = \int {\frac{2}{{x + 1}}dx} + \int {\frac{1}{{{x^2} + 2x + 2}}dx} \cr & {\text{completing the square}} \cr & = \int {\frac{2}{{x + 1}}dx} + \int {\frac{1}{{{x^2} + 2x + 1 + 1}}dx} \cr & = \int {\frac{2}{{x + 1}}dx} + \int {\frac{1}{{{{\left( {x + 1} \right)}^2} + 1}}dx} \cr & {\text{integrating}} \cr & = 2\ln \left| {x + 1} \right| + {\tan ^{ - 1}}\left( {x + 1} \right) + C \cr}