Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.5 Partial Fractions - 7.5 Exercises: 18

Answer

$$12\ln \left| {x - 4} \right| - 9\ln \left| {x + 3} \right| + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{21{x^2}}}{{{x^3} - {x^2} - 12x}}dx} \cr & = \int {\frac{{21{x^2}}}{{x\left( {{x^2} - x - 12} \right)}}dx} = \int {\frac{{21x}}{{{x^2} - x - 12}}dx} \cr & {\text{partial fraction decomposition}} \cr & \frac{{21x}}{{{x^2} - x - 12}} = \frac{{21x}}{{\left( {x - 4} \right)\left( {x + 3} \right)}} \cr & \frac{{21x}}{{\left( {x - 4} \right)\left( {x + 3} \right)}} = \frac{A}{{x - 4}} + \frac{B}{{x + 3}} \cr & 21x = A\left( {x + 3} \right) + B\left( {x - 4} \right) \cr & {\text{for }}x = 4 \cr & 21\left( 4 \right) = A\left( {4 + 3} \right) + B\left( {4 - 4} \right) \cr & 84 = 7A \cr & 12 = A \cr & {\text{for }}x = - 3 \cr & 21\left( { - 3} \right) = A\left( { - 3 + 3} \right) + B\left( { - 3 - 4} \right) \cr & - 63 = - 7B \cr & 9 = B \cr & {\text{substituting the values}} \cr & \frac{A}{{x - 4}} + \frac{B}{{x + 3}} = \frac{{12}}{{x - 4}} + \frac{9}{{x + 3}} \cr & \int {\frac{{21{x^2}}}{{{x^3} - {x^2} - 12x}}dx} = \int {\left( {\frac{{12}}{{x - 4}} + \frac{9}{{x + 3}}} \right)} dx \cr & {\text{sum rule}} \cr & = \int {\frac{{12}}{{x - 4}}} dx - \int {\frac{9}{{x + 3}}} dx \cr & {\text{integrating}} \cr & = 12\ln \left| {x - 4} \right| - 9\ln \left| {x + 3} \right| + C \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.