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

Sep 23, 2014

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

By decomposing into smaller fractions,

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

by taking the common denominator,

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

By comparing the numerators,

$A \left(3 x + 1\right) + B \left(x + 3\right) = 2 x$

by plugging in $x = - 3$,

$R i g h t a r r o w - 8 A = - 6 R i g h t a r r o w A = \frac{- 6}{- 8} = \frac{3}{4}$

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

$R i g h t a r r o w \frac{8}{3} B = - \frac{2}{3} R i g h t a r r o w B = - \frac{1}{4}$

Hence,

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