Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.4 - Integration of Rational Functions by Partial Fractions. - 7.4 Exercises - Page 501: 53

Answer

$$ \int \ln \left(x^{2}-x+2\right) d x = \left(x-\frac{1}{2}\right) \ln \left(x^{2}-x+2\right)-2 x+\sqrt{7} \tan ^{-1} \frac{2 x-1}{\sqrt{7}}+C $$ where $C$ be an arbitrary constant.

Work Step by Step

$$ \int \ln \left(x^{2}-x+2\right) d x $$ $$ \text { Let } u=\ln \left( x^{2}-x+2\right), \quad\quad d v=d x .\\ \text { Then } d u=\frac{2 x-1}{x^{2}-x+2} d x, v=x, $$ and (by integration by parts) , we have : \begin{aligned} \int \ln \left(x^{2}-x+2\right) d x &=x \ln \left(x^{2}-x+2\right)-\int \frac{2 x^{2}-x}{x^{2}-x+2} d x \\ & =x \ln \left(x^{2}-x+2\right)-\int\left(2+\frac{x-4}{x^{2}-x+2}\right) d x \\ &=x \ln \left(x^{2}-x+2\right)-2 x-\int \frac{\frac{1}{2}(2 x-1)}{x^{2}-x+2} d x + \\ & \quad\quad \quad\quad +\frac{7}{2} \int \frac{d x}{\left(x-\frac{1}{2}\right)^{2}+\frac{7}{4}} \\ &=x \ln \left(x^{2}-x+2\right)-2 x-\frac{1}{2} \ln \left(x^{2}-x+2\right)+ \\ & \quad\quad \quad\quad +\frac{7}{2} \int \frac{\frac{\sqrt{7}}{2} d u} {\frac{7}{4}\left(u^{2}+1\right)} \\ &\quad\quad \quad\quad \quad\quad \quad\left[\begin{array}{c} \text { where } x-\frac{1}{2}=\frac{\sqrt{7}}{2} u, \\ d x=\frac{\sqrt{7}}{2} d u, \\ \left(x-\frac{1}{2}\right)^{2}+\frac{7}{4}=\frac{7}{4}\left(u^{2}+1\right) \end{array}\right] \\ &=\left(x-\frac{1}{2}\right) \ln \left(x^{2}-x+2\right)-2 x+\sqrt{7} \tan ^{-1} u+C \\ &=\left(x-\frac{1}{2}\right) \ln \left(x^{2}-x+2\right)-2 x+\sqrt{7} \tan ^{-1} \frac{2 x-1}{\sqrt{7}}+C \end{aligned} So the integral: $$ \int \ln \left(x^{2}-x+2\right) d x = \left(x-\frac{1}{2}\right) \ln \left(x^{2}-x+2\right)-2 x+\sqrt{7} \tan ^{-1} \frac{2 x-1}{\sqrt{7}}+C $$ where $C$ be an arbitrary constant.
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