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

1 Answer
Sep 23, 2014

{2x}/{(x+3)(3x+1)}={3/4}/(x+3)-{1/4}/(3x+1)2x(x+3)(3x+1)=34x+3143x+1

By decomposing into smaller fractions,

{2x}/{(x+3)(3x+1)}=A/(x+3)+B/(3x+1)2x(x+3)(3x+1)=Ax+3+B3x+1

by taking the common denominator,

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

By comparing the numerators,

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

by plugging in x=-3x=3,

Rightarrow -8A=-6 Rightarrow A={-6}/{-8}=3/48A=6A=68=34

by plugging in x=-1/3x=13,

Rightarrow 8/3B=-2/3 Rightarrow B=-1/483B=23B=14

Hence,

{2x}/{(x+3)(3x+1)}={3/4}/(x+3)-{1/4}/(3x+1)2x(x+3)(3x+1)=34x+3143x+1