Write as a single fraction in its simplest form $\dfrac{3}{1-x^2}-\dfrac{2}{x+1}$. Answer ...

1. Factorise the denominator and identify a common denominator.

$1-x^2=(1-x)(1+x)=-(x-1)(x+1)$

So
$\frac{3}{1-x^2}-\frac{2}{x+1}=\frac{3}{-(x-1)(x+1)}-\frac{2}{x+1}.$

2. Write both terms over the same denominator and simplify the numerator.

Using common denominator $(x+1)(x-1)$,
$\frac{3}{1-x^2}-\frac{2}{x+1}=-\frac{3}{(x-1)(x+1)}-\frac{2(x-1)}{(x+1)(x-1)}$
$=\frac{-3-2(x-1)}{(x+1)(x-1)}$
$=\frac{-3-2x+2}{(x+1)(x-1)}$
$=\frac{-2x-1}{(x+1)(x-1)}.$

Equivalently, multiplying numerator and denominator by $-1$ gives
$\frac{2x+1}{(x+1)(1-x)}.$