Answer
See below
Work Step by Step
Given: $y'=(y-3)(y+1)$
Rewrite as: $\frac{dy}{dx}=(y-3)(y+1)\\\rightarrow \frac{dy}{(y-3)(y+1)}=dx\\
$
Integrate both sides:
$\int \frac{dy}{(y-3)(y+1)}=\int dx$
To solve this: $ \frac{1}{(y-3)(y+1)}=\frac{A}{y-3}+\frac{B}{y+1}=\frac{(A+B)y+(A-3B)}{(y-3)(y+1)}$
We have the system: $A+B=0\\A-3B=1$
then $A=-B\\
(-B)-3B=1\\ \rightarrow B=-\frac{1}{4}\\
\rightarrow A=\frac{1}{4}$
Hence, $\int \frac{dy}{(y-3)(y+1)}=\int\frac{dy}{4(y-3)}-\int\frac{dy}{4(y+1)}\\
=\frac{1}{4}\int\frac{dy}{y-3}-\frac{1}{4}\int \frac{dy}{y+1}\\
=\frac{1}{4}\ln(y-3)-\frac{1}{4}\ln (y+1)\\
=\frac{1}{4}(\ln(y-3)-\ln(y+1))\\
=\frac{1}{4}\ln(\frac{y-3}{y+1})$
Solve this, we get $\frac{1}{4}\ln(\frac{y-3}{y+1})=x+c$
where $c_1=0\\c_2=2\\c_3=4\\c_4=8$
