Since the degree of the numerator is #=# to the degree of the denominator, we must perform a long division
#x(x+1)^2=x(x^2+2x+1)=x^3+2x^2+x#
Now, we perform the long division
#color(white)(aaaa)##2x^3-x^2+ x+5##color(white)(aaaa)##∣##x^3+2x^2+x#
#color(white)(aaaa)##2x^3+4x^2+2x##color(white)(aaaaa)##∣##2#
#color(white)(aaaaaa)##0-5x^2-x+5#
Therefore,
#(2x^3-x^2+ x+5)/(x(x+1)^2)=2+(-5x^2-x+5)/(x(x+1)^2)#
We can make the partial fraction decomposition
#(-5x^2-x+5)/(x(x+1)^2)=A/(x)+B/(x+1)^2+C/(x+1)#
#=(A(x+1)^2+Bx+Cx(x+1))/((x(x+1)^2)#
So,
#-5x^2-x+5=A(x+1)^2+Bx+Cx(x+1)#
Let #x=0#, #=>#, #5=A#
Let #x=-1#, #=>#, #1=-B#, #=>#, #B=-1#
Coefficients of #x^2#, #=>#,#-5=A+C#, #=>#, #C=-10#
Finally we get
#(2x^3-x^2+ x+5)/(x(x+1)^2)=2+5/(x)-1/(x+1)^2-10/(x+1)#
