How do you express #(x+1)/( (x^2 )*(x-1) )# in partial fractions?

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Answer 1 · 2016-04-02T10:22:36

#(x+1)/(x^2(x-1))=-1/x^2-2/x+2/(x-1)#

From the given #(x+1)/(x^2(x-1))#, we set up the equations with the needed unknown constants represented by letters

#(x+1)/(x^2(x-1))=A/x^2+B/x+C/(x-1)#

The Least Common Denominator is #x^2(x-1)#.

Simplify

#(x+1)/(x^2(x-1))=A/x^2+B/x+C/(x-1)#

#(x+1)/(x^2(x-1))=(A(x-1)+B(x^2-x)+Cx^2)/(x^2(x-1))#

Expand the numerator of the right side of the equation

#(x+1)/(x^2(x-1))=(Ax-A+Bx^2-Bx+Cx^2)/(x^2(x-1))#

Rearrange from highest degree to the lowest degree the terms of the numerator at the right side

#(x+1)/(x^2(x-1))=(Bx^2+Cx^2+Ax-Bx-A)/(x^2(x-1))#

Now let us fix the numerators on both sides so that the numerical coefficients are matched.

#(0*x^2+1*x+1*x^0)/(x^2(x-1))=((B+C)x^2+(A-B)x-A*x^0)/(x^2(x-1)#

We can now have the equations to solve for A, B, C

#B+C=0#
#A-B=1#
#-A=1#

Simultaneous solution results to
#A=-1#
#B=-2#
#C=2#

We now have the final equivalent
#(x+1)/(x^2(x-1))=-1/x^2-2/x+2/(x-1)#

God bless....I hope the explanation is useful.