Hint: \(\int \frac{d x}{(x+1)(x-2)(x-3)}=\int\left(\frac{A}{x+1}+\frac{B}{(x-2)}+\frac{C}{x-3}\right) d x\)
\(\begin{aligned}
& \frac{1}{(x+1)(x-2)(x-3)}=\frac{A(x-2)(x-3)+B(x+1)(x-3)+C(x+1)(x-2)}{(x+1)(x-2)(x-3)} \\
& \Rightarrow 1=A(x-2)(x-3)+B(x+1)(x-3)+C(x+1)(x-2) \\
& \text { for } x=-1 \\
& 1=A(-3)(-4) \Rightarrow A=\frac{1}{12} \\
& \text { for } x=2 \\
& 1=B(3)(-1) \Rightarrow B=-\frac{1}{3}
\end{aligned}\)
\(\text { for } x=3\)
\(\begin{aligned} & 1=C(4)(1) \Rightarrow C=\frac{1}{4} \\ & \therefore I=\frac{1}{12} \int \frac{1}{x+1} d x-\frac{1}{3} \int \frac{1}{x-2} d x+\frac{1}{4} \int \frac{1}{x-3} d x \\ & =\frac{1}{12} \ln |x+1|-\frac{1}{3} \ln |x-2|+\frac{1}{4} \ln |x-3|+C \\ & =\frac{1}{12}\left[\ln |x+1|-\ln |x-2|^4+\ln |x-3|^3\right]+C \\ & =\frac{1}{12} \ln \frac{(|x+1|)(|x-3|)^3}{(x-2)^4}-C \\ & \therefore k=12\end{aligned}\)