Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

Chapter 7 - Integration Techniques - 7.5 Partial Fractions - 7.5 Exercises - Page 549: 43

Answer

$$\ln \left| {x + 1} \right| + {\tan ^{ - 1}}\left( x \right) + C$$

Work Step by Step

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

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