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

Question

Charles Arocha · Accepted Answer

#{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)#

Luke Alvarez · Answer

undefined To find the partial fraction decomposition of $ \frac{2x}{(x+3)(3x+1)} $, you first express the fraction in the form of partial fractions. Here's how you do it:

1. Write the fraction as $ \frac{A}{x+3} + \frac{B}{3x+1} $.
2. Multiply both sides by the denominator $(x+3)(3x+1)$ to get rid of the fractions.
3. Solve for $ A $ and $ B $ by equating coefficients of like terms on both sides.

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

Multiply both sides by $(x+3)(3x+1)$:

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

Expand the right side and equate coefficients:

$$ 2x = 3Ax + A + Bx + 3B $$

Now, equate coefficients of like terms:

For $ x $ terms: $ 2 = 3A + B $  
For constant terms: $ 0 = A + 3B $

Now, solve these two equations to find the values of $ A $ and $ B $, which will give you the partial fraction decomposition.