For the functions $f(x)=\frac{2}{x+3}$ and $g(x)=\frac{5}{x-1}$, find the composition $f \circ g$ and simplify your answer as much as possible. Write the domain using interval notation.
\[
(f \circ g)(x)=
\]
Domain of $f \circ g$ :
\begin{tabular}{|ccc|}
\hline$\frac{\square}{\square}$ & $\square^{\square}$ & $\sqrt{\square}$ \\
$\square$ 미미 & $(\square, \square)$ & {$[\square, \square]$} \\
$\square \cup \square$ & $(\square, \square]$ & {$[\square, \square)$} \\
$\varnothing$ & $\infty$ & $-\infty$ \\
$\times$ & & 6 \\
\hline
\end{tabular}

Step 1 ：Define the functions $f(x) = \frac{2}{x+3}$ and $g(x) = \frac{5}{x-1}$

Step 2 ：Find the composition of the functions $f$ and $g$ by substituting $g(x)$ into $f(x)$, which gives $(f \circ g)(x) = f(g(x)) = \frac{2}{3 + \frac{5}{x - 1}}$

Step 3 ：Simplify the composition function to get $(f \circ g)(x) = \frac{2(x - 1)}{3x + 2}$

Step 4 ：Find the domain of the composition function by setting the denominator equal to zero and solving for $x$, which gives $x = -\frac{2}{3}$

Step 5 ：Exclude $x = -\frac{2}{3}$ from the domain because it makes the denominator of the composition function equal to zero

Step 6 ：Also exclude $x = 1$ from the domain because it makes the denominator of the function $g(x)$ equal to zero

Step 7 ：Conclude that the domain of the composition function is all real numbers except $x = -\frac{2}{3}$ and $x = 1$, which can be written in interval notation as $(-\infty, -\frac{2}{3}) \cup (-\frac{2}{3}, 1) \cup (1, \infty)$

Step 8 ：\(\boxed{(f \circ g)(x) = \frac{2(x - 1)}{3x + 2}}\)

Step 9 ：\(\boxed{\text{Domain of } f \circ g = (-\infty, -\frac{2}{3}) \cup (-\frac{2}{3}, 1) \cup (1, \infty)}\)