Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 8 - Further Techniques and Applications of Integration - Chapter Review - Review Exercises - Page 455: 18

Answer

$$x\ln \left| {4x + 5} \right| - x + \frac{5}{4}\ln \left| {4x + 5} \right| + C$$

Work Step by Step

$$\eqalign{ & \int {\ln \left| {4x + 5} \right|} dx \cr & {\text{setting }}\,\,\,\,\,\,u = \ln \left| {4x + 5} \right|{\text{ then }}du = \frac{4}{{4x + 5}}dx\,\,\,\,\,\,\,\,\,\, \cr & {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = dx{\text{ then }}v = x \cr & {\text{Substituting these values into the formula for integration by parts}} \cr & \int u dv = uv - \int {vdu} \cr & \int {\ln \left| {4x + 5} \right|} dx = \ln \left| {4x + 5} \right|\left( x \right) - \int {\left( x \right)\left( {\frac{4}{{4x + 5}}dx} \right)} \cr & {\text{simplifying}} \cr & \int {\ln \left| {4x + 5} \right|} dx = x\ln \left| {4x + 5} \right| - \int {\frac{{4x}}{{4x + 5}}} dx \cr & {\text{use the long division for }}\frac{{4x}}{{4x + 5}}{\text{ to obtain }}1 - \frac{5}{{4x + 5}}.{\text{ then}} \cr & \int {\ln \left| {4x + 5} \right|} dx = x\ln \left| {4x + 5} \right| - \int {\left( {1 - \frac{5}{{4x + 5}}} \right)} dx \cr & \int {\ln \left| {4x + 5} \right|} dx = x\ln \left| {4x + 5} \right| - \int {dx} + \int {\frac{5}{{4x + 5}}} dx \cr & {\text{integrating}} \cr & \int {\ln \left| {4x + 5} \right|} dx = x\ln \left| {4x + 5} \right| - \int {dx} + \frac{5}{4}\int {\frac{4}{{4x + 5}}} dx3 \cr & \int {\ln \left| {4x + 5} \right|} dx = x\ln \left| {4x + 5} \right| - x + \frac{5}{4}\ln \left| {4x + 5} \right| + C \cr} $$
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