We are given:
\[
\int \frac{1}{\sqrt[5]{(x - 1)^4 (x + 3)^6}} \, dx = A \left( \frac{\alpha x - 1}{\beta x + 3} \right)^B + C
\]
Step 1: Simplify the integral
\[
I = \int \frac{1}{(x - 1)^{4/5} (x + 3)^{6/5}} \, dx
\]
\[
I = \int \frac{1}{(x - 1)^{4/5} (x + 3)^2} (x + 3)^{4/5} \, dx
\]
Step 2: Substitution
Let:
\[
t = \frac{x - 1}{x + 3}
\]
Then,
\[
dt = \frac{4}{(x + 3)^2} \, dx \quad \Rightarrow \quad \frac{dx}{(x + 3)^2} = \frac{dt}{4}
\]
Hence,
\[
I = \frac{1}{4} \int t^{-4/5} \, dt
\]
Step 3: Integration
\[
I = \frac{1}{4} \cdot \frac{t^{1/5}}{1/5} + C
\]
\[
I = \frac{5}{4} \left( \frac{x - 1}{x + 3} \right)^{1/5} + C
\]
Step 4: Comparing with given form
We can write:
\[
A = \frac{5}{4}, \quad \alpha = 1, \quad \beta = 1, \quad B = \frac{1}{5}
\]
Step 5: Calculating the required value
\[
\alpha + \beta + 20AB = 2 + 20 \times \frac{5}{4} \times \frac{1}{5}
\]
\[
= 2 + 20 \times \frac{1}{4} = 2 + 5 = 7
\]
Final Answer:
\[
\boxed{\alpha + \beta + 20AB = 7}
\]
Consider the given integral:
\[I = \int \frac{1}{(x - 1)^{4/5}(x + 3)^{6/5}} dx.\]
To simplify, let:
\[I = \int \frac{1}{(x - 1)^{4/5}(x + 3)^{6/5}} dx = \int \frac{1}{(x - 1)^{4/5}(x + 3)^2} dx.\]
Substitute:
\[t = \frac{x - 1}{x + 3} \implies dt = \frac{4}{(x + 3)^2} dx.\]
Thus, the integral becomes:
\[I = \frac{1}{4} \int t^{-4/5} dt.\]
Integrating:
\[I = \frac{1}{4} \cdot \frac{t^{1/5}}{1/5} + C = \frac{5}{4} t^{1/5} + C.\]
Substituting back \(t = \frac{x - 1}{x + 3}\):
\[I = \frac{5}{4} \left(\frac{x - 1}{x + 3}\right)^{1/5} + C.\]
Comparing with:
\[\int \frac{1}{\sqrt[5]{(x - 1)^4(x + 3)^6}} dx = A \left(\frac{\alpha x - 1}{\beta x + 3}\right)^B + C.\]
We have:
\[A = \frac{5}{4}, \quad \alpha = \beta = 1, \quad B = \frac{1}{5}.\]
Calculating \(\alpha + \beta + 20AB\):
\[\alpha + \beta + 20AB = 1 + 1 + 20 \times \frac{5}{4} \times \frac{1}{5} = 7.\]
Answer: 7.
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