x/(x+1)(x-2)
=a/(x+1)+b/(x-2)
=(ax-2a+bx+b)/(x+1)(x-2)
x=ax-2a+bx+b
所以a+b=1
-2a+b=0
a=1/3,b=2/3
原式=∫1/3*1/(x+1)dx+∫2/3*1/(x-2)dx
=1/3*ln|x+1|+2/3*ln|x-2|+C