(x+1)/(x^2(x-2)=(Ax+B)/x^2+(Cx+D)/(x-2)^2.
Therefore, Ax^3-4Ax^2+4Ax+Bx^2-4Bx+4B+Cx^3+Dx^2=x+1.
Equating terms: 4B=1 (constant term), so B=1/4; 4A-4B=1 (x term), so 4A=2 and A=1/2; -4A+B+D=0 (x^2 term), so D=4A-B=2-1/4=7/4; A+C=0 (x^3 term), so C=-1/2.
Therefore the partial fractions are: (2x+1)/4x^2+(7-2x)/4(x-2)^2.