How do you express (x+1)/( (x^2 )*(x-1) )x+1(x2)(x1) in partial fractions?

1 Answer

(x+1)/(x^2(x-1))=-1/x^2-2/x+2/(x-1)x+1x2(x1)=1x22x+2x1

Explanation:

From the given (x+1)/(x^2(x-1))x+1x2(x1), 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)x+1x2(x1)=Ax2+Bx+Cx1

The Least Common Denominator is x^2(x-1)x2(x1).

Simplify

(x+1)/(x^2(x-1))=A/x^2+B/x+C/(x-1)x+1x2(x1)=Ax2+Bx+Cx1

(x+1)/(x^2(x-1))=(A(x-1)+B(x^2-x)+Cx^2)/(x^2(x-1))x+1x2(x1)=A(x1)+B(x2x)+Cx2x2(x1)

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))x+1x2(x1)=AxA+Bx2Bx+Cx2x2(x1)

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))x+1x2(x1)=Bx2+Cx2+AxBxAx2(x1)

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)0x2+1x+1x0x2(x1)=(B+C)x2+(AB)xAx0x2(x1)

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

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

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

We now have the final equivalent
(x+1)/(x^2(x-1))=-1/x^2-2/x+2/(x-1)x+1x2(x1)=1x22x+2x1

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