Answer
$\frac{(x-5)}{4(x-3)}$.
Work Step by Step
The given expression is
$=\frac{x+5}{4x+12}+\frac{x}{9-x^2}$
First denominator $=4x+12$.
Factor $=4(x+3)$.
$=4(3+x)$
Second denominator $=9-x^2$.
Factor $=3^2-x^2$
Use the algebraic identity $a^2-b^2=(a+b)(a-b)$.
$=(3+x)(3-x)$
Substitute both factors into the given expression.
$=\frac{x+5}{4(3+x)}+\frac{x}{(3+x)(3-x)}$
The LCM of both denominators is $=4(3+x)(3-x)$.
$=\frac{x+5}{4(3+x)} \times \frac{3-x}{3-x}+\frac{x}{(3+x)(3-x)}\times \frac{4}{4}$
Simplify.
$=\frac{(x+5)(3-x)}{4(3+x)(3-x)}+\frac{4x}{4(3+x)(3-x)}$
$=\frac{(x+5)(3-x)+4x}{4(3+x)(3-x)}$
$=\frac{3x-5x-x^2+15+4x}{4(3+x)(3-x)}$
Simplify.
$=\frac{-x^2+2x+15}{4(3+x)(3-x)}$
Factor the numerator as shown below.
$=-x^2+2x+15$
$=-x^2+5x-3x+15$
$=-x(x-5)-3(x-5)$
$=(x-5)(-x-3)$
$=-(x-5)(x+3)$
Substitute into the fraction.
$=\frac{-(x-5)(x+3)}{4(3+x)(3-x)}$
Cancel common terms.
$=\frac{-(x-5)}{4(3-x)}$
Rearrange.
$=\frac{(x-5)}{4(x-3)}$.
