Step 1: Understanding the Concept:
The integrand contains a product of linear factors raised to fractional powers in the denominator.
Specifically, it is of the form \( (x-a)^{-m}(x-b)^{-n} \) where \( m + n = \frac{3}{4} + \frac{5}{4} = 2 \).
For integrals where the sum of powers equals 2, a substitution of the ratio \( t = \frac{x-a}{x-b} \) often reduces the expression to a basic power integral.
Key Formula or Approach:
1. Factor out a term to create a ratio.
2. Substitution: \( t = \frac{x-1}{x+2} \).
3. Derivative: \( \frac{dt}{dx} = \frac{(x+2)(1) - (x-1)(1)}{(x+2)^2} = \frac{3}{(x+2)^2} \).
Step 2: Detailed Explanation:
Let \( I = \int \frac{1}{(x-1)^{3/4}(x+2)^{5/4}} dx \).
Multiply and divide the denominator by \( (x+2)^{3/4} \):
\[ I = \int \frac{1}{\left(\frac{x-1}{x+2}\right)^{3/4} (x+2)^{3/4} (x+2)^{5/4}} dx \]
\[ I = \int \frac{1}{\left(\frac{x-1}{x+2}\right)^{3/4} (x+2)^2} dx \]
Now, let \( t = \frac{x-1}{x+2} \).
Differentiating both sides: \( dt = \frac{3}{(x+2)^2} dx \), which means \( \frac{dx}{(x+2)^2} = \frac{dt}{3} \).
Substitute these into the integral:
\[ I = \int \frac{1}{t^{3/4}} \cdot \frac{dt}{3} = \frac{1}{3} \int t^{-3/4} dt \]
Evaluate the integral using the power rule:
\[ I = \frac{1}{3} \left[ \frac{t^{-3/4 + 1}}{-3/4 + 1} \right] + C = \frac{1}{3} \left[ \frac{t^{1/4}}{1/4} \right] + C \]
\[ I = \frac{1}{3} \cdot 4 \cdot t^{1/4} + C = \frac{4}{3} t^{1/4} + C \]
Substitute back for \( t \):
\[ I = \frac{4}{3} \left( \frac{x-1}{x+2} \right)^{1/4} + C \]
Step 3: Final Answer:
The result of the integration is \( \frac{4}{3}\left(\frac{x-1}{x+2}\right)^{1/4}+C \).
This matches Option (C).