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: 29

Answer

$$-\ln |x|+\ln |x-1|+\frac{1}{(x-1)}-\frac{1}{2(x-1)^{2}}+C$$

Work Step by Step

Given $$\int \frac{d x}{x(x-1)^{3}}$$ Since \begin{aligned} \frac{1}{x(x-1)^{3}} &=\frac{A}{x}+\frac{B}{(x-1)}+\frac{C}{(x-1)^{2}}+\frac{D}{(x-1)^{3}} \\ &=\frac{A(x-1)^{3}+B x(x-1)^{2}+C x(x-1)+D x}{x(x-1)^{3}} \\ 1 &=A(x-1)^{3}+B x(x-1)^{2}+C x(x-1)+D x \end{aligned} Then \begin{align*} \text{at } x&=0 \ \ \ \ \to A=-1\\ \text{at } x&=1\ \ \ \ \to D=1\\ \text{at } x&= -1\ \ \ \ \to -3=-2 B+C\\ \text{at } x&=2\ \ \ \ \to 0= B+C \end{align*} Hence $B=1,\ \ \ C=-1$ and \begin{aligned} \int \frac{d x}{x(x-1)^{3}} &=\int \frac{-d x}{x}+\int \frac{d x}{(x-1)}-\int \frac{d x}{(x-1)^{2}}+\int \frac{d x}{(x-1)^{3}} \\ &=-\ln |x|+\ln |x-1|+\frac{1}{(x-1)}-\frac{1}{2(x-1)^{2}}+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.