How do I find the partial fraction decomposition of $\frac{2 x}{(x + 3) (3 x + 1)}$ $(2x)/((x+3)(3x+1))$ ?

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Answer 1 · 2014-09-23T00:05:34

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

By decomposing into smaller fractions,

$\frac{2 x}{(x + 3) (3 x + 1)} = \frac{A}{x + 3} + \frac{B}{3 x + 1}$ ${2x}/{(x+3)(3x+1)}=A/(x+3)+B/(3x+1)$

by taking the common denominator,

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

By comparing the numerators,

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

by plugging in $x = - 3$ $x=-3$ ,

$\Rightarrow - 8 A = - 6 \Rightarrow A = \frac{- 6}{- 8} = \frac{3}{4}$ $Rightarrow -8A=-6 Rightarrow A={-6}/{-8}=3/4$

by plugging in $x = - \frac{1}{3}$ $x=-1/3$ ,

$\Rightarrow \frac{8}{3} B = - \frac{2}{3} \Rightarrow B = - \frac{1}{4}$ $Rightarrow 8/3B=-2/3 Rightarrow B=-1/4$

Hence,

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

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