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,

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

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