不定积分求法,∫(2x+1)/x(x-1)^2dx,写出具体步骤。

则 A(x-1)²+(Bx+C)x=2x+1

化简得 (A+B)x²+(C-2A)x+A=2x+1

∴A+B=0

C-2A=2

A=1

解得 A=1,B=-1,C=4

∴(2x+1)/[x(x-1)²]=1/x+(4-x)/(x-1)²=1/x-(x-1-3)/(x-1)²=1/x-1/(x-1)+3/(x-1)²

∫(2x+1)/[x(x-1)²] dx

=∫ [1/x-1/(x-1)+3/(x-1)²] dx

=∫ (1/x) dx-∫[1/(x-1)]dx+3∫1/(x-1)²dx

=ln|x|-ln|x-1|-3/(x-1)+C

=ln|x/(x-1)|-3/(x-1)+C

C为任意常数