Using partial fraction decomposition , break up the fraction into three different fractions added together:
#int frac{1}{(x-1)(x+2)(x-3)}dx#
#=int (frac{color(blue)(A)}{x-1}+frac{color(red)(B)}{x+2}+frac{color(green)(C)}{x-3})dx#
#---#
In order to find the values of #color(blue)(A), color(red)(B), and color(green)(C)#, use common denominators, and set it equal to the original fraction.
#color(blue)(A)(x+2)(x-3)+color(red)(B)(x-1)(x-3)+color(green)(C)(x-1)(x+2)=1#
Plug in any value for #x#. The easiest numbers to plug in are those which make one of the factors zero, because many terms cancel:
#"If x=1: "color(blue)(A)(1+2)(1-3)=1#
#color(blue)(A)=-1/6#
#"If x=-2: "color(red)(B)(-2-1)(-2-3)=1#
#color(red)(B)=1/15#
#"If x=3: "color(green)(C)(3-1)(3+2)=1#
#color(green)(C)=1/10#
#---#
The integral after plugging in values of #color(blue)(A), color(red)(B), and color(green)(C)# becomes:
#=int (frac{color(blue)(-1/6)}{x-1}+frac{color(red)(1/15)}{x+2}+frac{color(green)(1/10)}{x-3})dx#
#=-1/6ln|x-1|+1/15ln|x+2|+1/10ln|x-3|+C#
