Answer
$\{4-\sqrt 5,4+\sqrt 5\}$
Work Step by Step
First we determine the domain of the equation:
$$(-\infty,-1)\cup(-1,\infty).$$
Multiply each side by $4(x+1)$ to clear denominators:
$$\begin{align*}
4(x+1)\left(\dfrac{5}{x+1}+\dfrac{x-1}{4}\right)&=4(x+1)(2)\\
20+(x^2-1)&=8(x+1)\\
x^2+19&=8x+8.
\end{align*}$$
Bring the equation to the standard form:
$$\begin{align*}
x^2+19-8x-8&=0\\
x^2-8x+11&=0.
\end{align*}$$
Build a perfect square, then use the difference of squares to find the solutions:
$$\begin{align*}
(x^2-8x+16)-5&=0\\
(x-4)^2-5&=0\\
(x-4-\sqrt 5)(x-4+\sqrt 5)&=0\\
x-4-\sqrt 5=0&\text{ or }x-4+\sqrt 5=0\\
x=4+\sqrt 5&\text{ or }x=4-\sqrt 5.
\end{align*}$$
Both solutions are in the domain, so the solution set of the equation is:
$$\{4-\sqrt 5,4+\sqrt 5\}.$$
