Answer
$$\frac{1}{5} \ln |x-3|+\frac{1}{20} \ln |x+2|-\frac{1}{4} \ln |x-2|+C$$
Work Step by Step
Given $$\int \frac{d x}{(x-2)(x-3)(x+2)}$$
Since
\begin{align*}
\frac{1}{(x-2)(x-3)(x+2)}&=\frac{A}{(x-2)}+\frac{B}{(x-3)}+\frac{C}{(x+2)}\\
&=\frac{A(x-3)(x+2)+B(x-2)(x+2)+C(x-2)(x-3)}{(x-2)(x-3)(x+2)}\\
1&=A(x-3)(x+2)+B(x-2)(x+2)+C(x-2)(x-3)
\end{align*}
Then
\begin{align*}
\text{at } x&=2\ \ \ \ \ A=\frac{-1}{4} \\
\text{at } x&=3\ \ \ \ \ B=\frac{1}{20}\\
\text{at } x&=-2\ \ \ \ \ B=\frac{1}{5}
\end{align*}
Hence
\begin{align*}
\int \frac{1}{(x-2)(x-3)(x+2)}dx&=\frac{-1}{4}\int\frac{1}{(x-2)}dx+\frac{1}{20}\int\frac{1}{(x-3)}dx+\frac{1}{5}\int\frac{1}{(x+2)}dx\\
&=\frac{1}{5} \ln |x-3|+\frac{1}{20} \ln |x+2|-\frac{1}{4} \ln |x-2|+C
\end{align*}
