Answer
$x = \frac{4}{3}$
Work Step by Step
First, factor all expressions to their simplest forms:
$\dfrac{3}{x + 1} = \dfrac{1}{(x + 1)(x - 1)}$
Multiply each side of the equation by the LCD, which is $(x + 1)(x - 1)$, to eliminate the fractions:
$$\begin{align*}
(x+1)(x-1) \cdot \frac{3}{x+1}&=\frac{1}{(x+1)(x-1)} \cdot (x+1)(x-1)\\
\end{align*}$$
Simplify:
$$\require{cancel}
\begin{align*}
\cancel{(x+1)}(x-1) \cdot \frac{3}{\cancel{x+1}}&=\frac{1}{\cancel{(x+1)(x-1)}} \cdot \cancel{(x+1)(x-1)}\\
\\(x-1)(3)&=1\\
3x-3&=1\\
3x&=1+3\\
3x&=4\\
x&=\frac{4}{3}
\end{align*}$$
To check the solution, plug in the value we just found for $x$ into the original equation:
$$\begin{align*}
\frac{3}{\frac{4}{3} + 1} &\stackrel{?}{=} \frac{1}{\left(\frac{4}{3}\right)^2 - 1}\\
\frac{3}{\frac{7}{3}}&\stackrel{?}{=}\frac{1}{\frac{16}{9}-1}\\\
3\cdot \frac{3}{7}&\stackrel{?}{=}\frac{1}{\frac{16}{9}-\frac{9}{9}}\\
\frac{9}{7}&\stackrel{?}{=}\frac{1}{\frac{7}{9}}\\
\frac{9}{7}&\stackrel{\checkmark}{=}\frac{9}{7}
\end{align*}$$
