Question

If \[ \int \frac{1}{\sqrt[5]{(x - 1)^4}(x + 3)^6} \, dx = A \left( \frac{\alpha x - 1}{\beta x + 3} \right)^B + C, \] where \(C\) is the constant of integration, then the value of \(\alpha + \beta + 20AB\) is _______.

Answer 1

Correct Answer: 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.