If `int4(3-2x)^(-2)((3-2x)/(3+2x))^((1)/(3))dx=(3)/(alpha)((3+2x)/(3-2x))^((beta)/(gamma))+c`
`(beta and gamma" are prime nos.")`, then

Options

A. `alpha, beta, gamma` are in G.P.
B. `alpha, beta, gamma` are in H.P.
C. `alphaa,beta, gamma` are in A.P.
D. `alpha=beta gamma`

Answer

To solve the given integral equation, we need to find the values of \( \alpha \), \( \beta \), and \( \gamma \) under the conditions that \( \beta \) and \( \gamma \) are prime numbers. Let's break down the solution step by step.

### Step 1: Understand the Integral
We are given the integral:
\[ \int 4(3-2x)^{-2}\left(\frac{3-2x}{3+2x}\right)^{\frac{1}{3}}dx = \frac{3}{\alpha}\left(\frac{3+2x}{3-2x}\right)^{\frac{\beta}{\gamma}} + c \]
Here, we need to find the values of \( \alpha \), \( \beta \), and \( \gamma \).

### Step 2: Differentiate the Right Side
To find the relationship between \( \alpha \), \( \beta \), and \( \gamma \), we differentiate the right-hand side:
\[ \frac{d}{dx}\left(\frac{3}{\alpha}\left(\frac{3+2x}{3-2x}\right)^{\frac{\beta}{\gamma}}\right) \]
Using the chain rule:
\[ = \frac{3}{\alpha} \cdot \frac{\beta}{\gamma} \left(\frac{3+2x}{3-2x}\right)^{\frac{\beta}{\gamma}-1} \cdot \frac{d}{dx}\left(\frac{3+2x}{3-2x}\right) \]

### Step 3: Differentiate the Fraction
Now we differentiate \( \frac{3+2x}{3-2x} \):
\[ \frac{d}{dx}\left(\frac{3+2x}{3-2x}\right) = \frac{(3-2x)(2) - (3+2x)(-2)}{(3-2x)^2} \]
Simplifying this gives:
\[ = \frac{6 - 4x + 6 + 4x}{(3-2x)^2} = \frac{12}{(3-2x)^2} \]

### Step 4: Substitute Back into the Derivative
Substituting this back into our derivative gives:
\[ \frac{d}{dx}\left(\frac{3}{\alpha}\left(\frac{3+2x}{3-2x}\right)^{\frac{\beta}{\gamma}}\right) = \frac{3}{\alpha} \cdot \frac{\beta}{\gamma} \left(\frac{3+2x}{3-2x}\right)^{\frac{\beta}{\gamma}-1} \cdot \frac{12}{(3-2x)^2} \]

### Step 5: Set the Derivatives Equal
Now we set this equal to the left-hand side of the original equation:
\[ 4(3-2x)^{-2}\left(\frac{3-2x}{3+2x}\right)^{\frac{1}{3}} \]

### Step 6: Equate Coefficients
From the equation, we can equate coefficients:
1. The coefficient of \( (3-2x)^{-2} \):
\[ \frac{3 \cdot \beta \cdot 12}{\alpha \cdot \gamma} = 4 \]
Simplifying gives:
\[ 9\beta = \alpha \gamma \]

2. The powers of the base:
\[ \frac{\beta}{\gamma} - 1 = \frac{1}{3} \]
This simplifies to:
\[ \frac{\beta}{\gamma} = \frac{4}{3} \]

### Step 7: Find Prime Numbers
Since \( \beta \) and \( \gamma \) are prime numbers:
Let \( \beta = 4k \) and \( \gamma = 3k \). The only primes satisfying \( \beta \) and \( \gamma \) being prime are \( \beta = 2 \) and \( \gamma = 3 \).

### Step 8: Solve for Alpha
Substituting \( \beta \) and \( \gamma \) back into \( 9\beta = \alpha \gamma \):
\[ 9 \cdot 2 = \alpha \cdot 3 \]
This gives:
\[ \alpha = 6 \]

### Final Values
Thus, we find:
- \( \alpha = 6 \)
- \( \beta = 2 \)
- \( \gamma = 3 \)

### Conclusion
The final answer is:
- \( \alpha = 6 \)
- \( \beta = 2 \)
- \( \gamma = 3 \)