Answer
$$-\frac{1}{2}\ln{|x-2|}+\frac{1}{2}\ln{|x-4|}+C$$
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:
$\int\frac{dx}{(x-2)(x-4)}$
$=-\frac{1}{2}\int\frac{1}{x-2}dx+\frac{1}{2}\int\frac{1}{x-4}dx$
$=-\frac{1}{2}\ln{|x-2|}+\frac{1}{2}\ln{|x-4|}+C$
