How do you express as a partial fraction #x/((x+1)(x^2 -1))#?

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Answer 1 · 2015-08-22T02:58:48

#(-1/4)/(x+1) + (1/2)/(x+1)^2 + (1/4)/(x-1) = 1/4[(-1)/(x+1) + (2)/(x+1)^2 + 1/(x-1)]#

First we need to finish factoring the denominator into factors that are irreducible using Real coefficients:

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

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

So we need:

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

Clear the denominator (multiply by #((x+1)^2(x-1))#on both sides), to get:

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

#Ax^2-A + Bx -B +Cx^2+2Cx+C = x#

#Ax^2+Cx^2 +Bx +2Cx -A-B+C = 0x^2 +1x+0#

So we need to solve:

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

From the first equation, we get: #A = -C# and we can substitue in the third equation to get

Eq 2: #B+2C = 1#
and: #-B+2C = 0#

Adding gets us #C = 1/4#, so #A = -1/4# and subtracting gets us #B = 1/2#

#A/(x+1) + B/(x+1)^2 + C/(x-1) = (-1/4)/(x+1) + (1/2)/(x+1)^2 + (1/4)/(x-1) #

# = 1/4[(-1)/(x+1) + (2)/(x+1)^2 + 1/(x-1)]#