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Answer 1 · 2018-01-01T15:12:05

The answer is #=(1/8)/(x)+(1/4)/(x+2)+(-3/8)/(x+4)#

I assume that #(x+1)# is the numerator

Perform the decomposition into partial fractions

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

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

The denominators are the same, compare the numerators

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

Let #x=0#, #=>#, #1=8A#, #=>#, #A=1/8#

Let #x=-2#, #=>#, #-1=-4B#, #=>#, #B=1/4#

Let #x=-4#, #=>#, #-3=3C#, #=>#, #C=-3/8#

So,

#(x+1)/((x)(x+2)(x+4))=(1/8)/(x)+(1/4)/(x+2)+(-3/8)/(x+4)#

Answer 2 · 2018-01-01T15:14:43

#(x+1)/((x)(x+2)(x+4))=1/(8x)+1/(4(x+2))-3/(8(x+4))#

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

Multiply both sides by #(x)(x+2)(x+4)#

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

#x=0; -2; -4# are roots which means we can substitute x and find A, B and C.

#x=0#
#1=8A+0B+0C=>A=1/8#

#x=-2#
#-1=-4B=>B=1/4#

#x=-4#
#-3=8C=>C=-3/8#

#(x+1)/((x)(x+2)(x+4))=1/(8x)+1/(4(x+2))-3/(8(x+4))#

(Method used above works only when we don't have multiple root. otherwise we would have to make a matrix)