How do I find the partial fraction decomposition of #(2x)/((x+3)(3x+1))# ?

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Answer 1 · 2014-09-23T00:05:34

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

By decomposing into smaller fractions,

#{2x}/{(x+3)(3x+1)}=A/(x+3)+B/(3x+1)#

by taking the common denominator,

#={A(3x+1)+B(x+3)}/{(x+3)(3x+1)}#

By comparing the numerators,

#A(3x+1)+B(x+3)=2x#

by plugging in #x=-3#,

#Rightarrow -8A=-6 Rightarrow A={-6}/{-8}=3/4#

by plugging in #x=-1/3#,

#Rightarrow 8/3B=-2/3 Rightarrow B=-1/4#

Hence,

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