# Chapter 8 - Techniques of Integration - 8.5 The Method of Partial Fractions - Exercises - Page 423: 9

$$-\frac{1}{2}\ln{|x-2|}+\frac{1}{2}\ln{|x-4|}+C$$

#### Work Step by Step

$\int$$\frac{dx}{(x-2)(x-4)}$ Writing $\frac{1}{(x-2)(x-4)}$ as partial fractions: $\frac{1}{(x-2)(x-4)} = \frac{A}{x-2}+\frac{B}{x-4}$, where A and B are real integer constants. Multiplying both sides by $(x-2)(x-4)$: $1=A(x-4)+B(x-2)$ If we let $x=2$, we can eliminate constant B to find constant A: $1=A(2-4)+B(2-2)$; and thus $1=-2A$, so $A=-\frac{1}{2}$. Similarly, if we let $x=4$, we can eliminate constant A to find constant B: $1=A(4-4)+B(4-2)$; and thus $1=2B$, so $B=\frac{1}{2}$. Thus, $\frac{1}{(x-2)(x-4)}$ could be integrated as partial fractions: $\int\frac{dx}{(x-2)(x-4)}$ $=-\frac{1}{2}\int\frac{1}{x-2}dx+\frac{1}{2}\int\frac{1}{x-4}dx$ $=-\frac{1}{2}\ln{|x-2|}+\frac{1}{2}\ln{|x-4|}+C$

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