Answer
$\dfrac{x^2 + 4x - 3}{(x + 1)(x - 1)}$
Restriction: $x \ne -1, 1$
Work Step by Step
Before performing the addition, find the least common denominator (LCD) of the two fractions. The LCD is $(x-1)(x+1)$. Convert the original fractions to equivalent fractions using the LCD:
$\dfrac{3(x - 1)}{(x + 1)(x - 1)} + \dfrac{x(x + 1)}{(x - 1)(x + 1)}$
Multiply to simplify:
$=\dfrac{3x - 3}{(x + 1)(x - 1)} + \dfrac{x^2 + x}{(x - 1)(x + 1)}$
Add the fractions:
$=\dfrac{3x - 3 + x^2 + x}{(x + 1)(x - 1)}$
Combine like terms:
$=\dfrac{x^2 + 4x - 3}{(x + 1)(x - 1)}$
Restrictions on $x$ occur when the value of $x$ makes the fraction undefined, which means that the denominator becomes $0$.
Set the denominators equal to $0$ to find restrictions:
First denominator:
$x + 1 = 0$
$x = -1$
Second denominator:
$x - 1 = 0$
$x = 1$
Thus, the restrictions are: $x \ne -1, 1$
