求不定积分1/((x-2)^2*(x-3))

Question

求不定积分1/((x-2)^2*(x-3))

Answer 1

∵1/[(x-2)²(x-3)]=-1/(x-2)-1/(x-2)²+1/(x-3)
提示：
设1/[(x-2)²(x-3)]=a/(x-2)+b/(x-2)²+c/(x-3)=[a(x-2)(x-3)+b(x-3)+c(x-2)²]/[(x-2)²(x-3)]=[(a+c)x²+(-5a+b-4c)x+(6a-3b+4c)]/[(x-2)²(x-3)]
比系数得：a+c=0
-5a+b+4c=0
6a-3b+4c=1
解得a=-1,b=-1,c=1
∴∫1/[(x-2)²(x-3)] dx
=∫[-1/(x-2)-1/(x-2)²+1/(x-3)]dx
=-∫1/(x-2)-∫1/(x-2)²dx+∫1/(x-3)dx
=-ln|x-2|+1/(x-2)+ln|x-3|+C
=ln|(x-3)/(x-2)|+1/(x-2)+C