Answer
$ \displaystyle \frac{1}{x+1}+\frac{1}{x-3}-\frac{2}{2x+9}$
Work Step by Step
Apply the Product and Quotient Rules for logarithms,
$r(x)=\ln|x+1|+\ln|x-3\}-\ln|-2x-9|$
... $|-2x-9|=|-(2x-9)|=|-2x-9|$
$r(x)=\ln|x+1|+\ln|x-3\}-\ln|2x+9|$
Apply $\displaystyle \frac{d}{dx}[\ln|u|]=\frac{1}{u}\cdot\frac{du}{dx}$
$r^{\prime}(x)=\displaystyle \frac{1}{x+1}+\frac{1}{x-3}-\frac{2}{2x+9}$
