Answer
The answer is $\dfrac{(x+3)(x^2-x-15)}{x(4x^2+5x+3)}$.
Work Step by Step
The simplification of the numerator can be done as follows:
$\dfrac{2x+5}{x}-\dfrac{x}{x-3}$
$=\dfrac{(2x+5)(x-3)-x^2}{x(x-3)}$ [ Take LCM of the rationals ]
$=\dfrac{2x^2-6x+5x-15-x^2}{x(x-3)}$ [ Expand the product ]
$=\dfrac{x^2-x-15}{x(x-3)}$ [ Simplify by adding/subtracting like terms as needed]
The simplification of the denominator can be done as follows:
$\dfrac{x^2}{x-3}-\dfrac{(x+1)^2}{x+3}$
$=\dfrac{x^2(x+3)-(x-3)(x+1)^2}{(x-3)(x+3)}$ [ Take LCM of rationals]
$=\dfrac{x^3+3x^2-(x-3)(x^2+2x+1)}{(x-3)(x+3)}$ [Expand and simplify]
$=\dfrac{x^3+3x^2-x^2(x-3)-2x(x-3)-(x-3)}{(x-3)(x+3)}$
$=\dfrac{x^3+3x^2-x^3+3x^2-2x^2+6x-x+3}{(x-3)(x+3)}$ [ Add/ subtract like terms]
$=\dfrac{4x^2+5x+3}{(x-3)(x+3)}$
Thus, the given expression can be simplified as follows:
$\dfrac{\dfrac{2x+5}{x}-\dfrac{x}{x-3}}{\dfrac{x^2}{x-3}-\dfrac{(x+1)^2}{x+3}}$
$=\dfrac{\dfrac{x^2-x-15}{x(x-3)}}{\dfrac{4x^2+5x+3}{(x-3)(x+3)}}$
$=\dfrac{(x^2-x-15)(x-3)(x+3)}{x(4x^2+5x+3)(x-3)}$ [Reciprocate and multiply]
$=\dfrac{(x^2-x-15)(x+3)}{x(4x^2+5x+3)}$ [Cancel common factor]
Therefore, the simplified expression is $\dfrac{(x^2-x-15)(x+3)}{x(4x^2+5x+3)}$.
