Calculus (3rd Edition)

Published by W. H. Freeman

Chapter 8 - Techniques of Integration - 8.6 Strategies for Integration - Exercises - Page 431: 12

Answer

$$\ln x-\ln |x-1|-\frac{1}{x-1}+C$$

Work Step by Step

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

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