$$\begin{aligned} & I=\int \frac{1}{\sqrt[5]{(x-1)^4(x+3)^6}} d x \\ & \frac{x+3}{x-1}=t \Rightarrow d x=\frac{-4}{(t-1)^2} d t \Rightarrow x=\left(\frac{3+t}{t-1}\right) \\ & \Rightarrow(x-1)^4(x+3)^6=(x-1)^5(x+3)^5\left(\frac{x+3}{x-1}\right) \\ & I=\int \frac{\frac{-4}{(t-1)^2} d t}{t^{1 / 5}\left(\frac{3+t}{t-1}-1\right)\left(\frac{3+t}{t-3}+3\right)} \\ & I=\int \frac{-4 d t}{t^{1 / 5}(16 t)}=\frac{5}{4}\left(\frac{x-1}{x+3}\right)^{1 / 5}+c \end{aligned}$$

Comparing,

$$\begin{aligned} & \Rightarrow A=\frac{5}{4}, B=\frac{1}{5}, \alpha=1, \beta=1 \\ & \Rightarrow \alpha+\beta+20 A B=1+1+20 \times \frac{5}{4} \times \frac{1}{5}=7 \end{aligned}$$