Question

The integral \(\int \frac{1}{\sqrt[4]{(x - 1)^3 (x + 2)^5}} \, dx\) is equal to :
(where C is a constant of integration)

• A. \(\frac{4}{3} \left( \frac{x - 1}{x + 2} \right)^{\frac{1}{4}} + C\)
• B. \(\frac{3}{4} \left( \frac{x + 2}{x - 1} \right)^{\frac{3}{4}} + C\)
• C. \(\frac{4}{3} \left( \frac{x - 1}{x + 2} \right)^{\frac{5}{4}} + C\)
• D. \(\frac{3}{4} \left( \frac{x + 2}{x - 1} \right)^{\frac{5}{4}} + C\)

Hint:

For integrals of the form \(\int (x - a)^{-m} (x - b)^{-n} \, dx\), if \(m + n\) is an integer, try the substitution \(t = \frac{x - a}{x - b}\). This usually simplifies the problem to a standard power rule integral.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is an integral involving a rational power of a quotient. A common substitution for integrals of the form \(\int \frac{1}{(x - a)^m (x - b)^n} \, dx\) where \(m + n = 2\) is \(t = \frac{x - a}{x - b}\).
Step 2: Detailed Explanation:
The integral is:
\[ I = \int \frac{dx}{(x - 1)^{3/4} (x + 2)^{5/4}} \]
Rewrite the denominator by dividing and multiplying by \((x + 2)^{3/4}\):
\[ I = \int \frac{dx}{\frac{(x - 1)^{3/4}}{(x + 2)^{3/4}} \cdot (x + 2)^{3/4} \cdot (x + 2)^{5/4}} = \int \frac{1}{\left( \frac{x - 1}{x + 2} \right)^{3/4}} \cdot \frac{1}{(x + 2)^2} \, dx \]
Let \(t = \frac{x - 1}{x + 2}\).
Differentiate w.r.t \(x\):
\[ dt = \frac{(x + 2)(1) - (x - 1)(1)}{(x + 2)^2} \, dx = \frac{3}{(x + 2)^2} \, dx \implies \frac{dx}{(x + 2)^2} = \frac{dt}{3} \]
Substituting into the integral:
\[ I = \frac{1}{3} \int t^{-3/4} \, dt = \frac{1}{3} \cdot \frac{t^{-3/4 + 1}}{-3/4 + 1} + C = \frac{1}{3} \cdot \frac{t^{1/4}}{1/4} + C = \frac{4}{3} t^{1/4} + C \]
Substitute back \(t = \frac{x - 1}{x + 2}\):
\[ I = \frac{4}{3} \left( \frac{x - 1}{x + 2} \right)^{\frac{1}{4}} + C \]
Step 3: Final Answer:
The integral is \(\frac{4}{3} \left( \frac{x - 1}{x + 2} \right)^{1/4} + C\), which is option (A).