How do you express #(x^2)/(x+1)^3# in partial fractions?

1 Answer

#x^2/(x+1)^3=1/(x+1)^3-2/(x+1)^2+1/(x+1)#

Explanation:

Set up the equation first by assigning letter variables A, B, C.

#x^2/(x+1)^3=A/(x+1)^3+B/(x+1)^2+C/(x+1)#

#x^2/(x+1)^3=A/(x+1)^3+(B(x+1))/(x+1)^3+(C(x+1)^2)/(x+1)^3#

#x^2/(x+1)^3=(A+B(x+1)+C(x+1)^2)/(x+1)^3#

#x^2/(x+1)^3=(A+Bx+B+Cx^2+2Cx+C)/(x+1)^3#

#(x^2+0*x+0*x^0)/(x+1)^3=(Cx^2+Bx+2Cx+(A+B+C)x^0)/(x+1)^3#

Set up the equations to solve for the values of A, B, C by equating the coefficients of each term

#Cx^2=1*x^2#

#Bx+2Cx=0*x#

#(A+B+C)x^0=0*x^0#

Therefore, after simplification, the equations are:

#C=1#
#B+2C=0#
#A=B+C=0#

Solving simultaneously, the values are

#A=1# and #B=-2# and #C=1#

so that

#x^2/(x+1)^3=1/(x+1)^3-2/(x+1)^2+1/(x+1)#

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