设
A/2x+B/(x-1)=[A(x-1)+B*2x]/2x(x-1)
=(Ax-A+2Bx)/2x(x-1)
=[(A+2B)x-A]/2x(x-1)
=1/2x(x-1)
则有
-A=1
A=-1
A+2B=0
2B=-A=1
B=1/2
所以
1/2x(x-1)=1/2(x-1)-1/2x
所以
∫dx/2x(x-1)
=∫dx/2(x-1)-∫dx/2x
=1/2∫d(x-1)/(x-1)-1/2∫dx/x
=1/2(ln|x-1|-ln|x|)
=1/2ln|(x-1)|/|x|