Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.5 The Method of Partial Fractions - Exercises - Page 423: 31


$$x+\ln |x|-3 \ln |x+1|+C$$

Work Step by Step

Given $$\int \frac{\left(x^{2}-x+1\right) d x}{x^{2}+x}$$ By using long division, we get: \begin{aligned} \frac{x^{2}-x+1}{x^{2}+x}&=1+\frac{-2 x+1}{x^{2}+x}\\ &=1+\frac{-2 x+1}{x(x+1)} \end{aligned} Then \begin{align*} \frac{-2 x+1}{x(x+1)}&=\frac{A}{x}+\frac{B}{x+1}\\ &=\frac{A(x+1)+B x}{x(x+1)}\\ -2x+1&=A(x+1)+B x \end{align*} \begin{align*} \text{at } x&=0 \ \ \ \ \to A=1 \\ \text{at } x&= -1\ \ \ \ \to B=-3 \end{align*} Hence \begin{aligned} \int \frac{x^{2}-x+1}{x^{2}+x} d x &=\int 1 \, d x+\int \frac{1}{x} d x-\int \frac{3}{x+1} d x \\ &=x+\ln |x|-3 \ln |x+1|+C \end{aligned}
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.