`f'(dx)/((x+6)^(8//7)(x-8)^(6//7))` is equal to

Options

A. `((x+6)/(x-8))^(1/7)+e`
B. `((x-8)/(x+6))^(1/7)+e`
C. `1/2((x-8)/(x+6))^(1/7)+e`
D. None of these

Explanation

To solve the integral \( I = \int \frac{dx}{(x+6)^{\frac{8}{7}}(x-8)^{\frac{6}{7}}} \), we will follow these steps:

### Step 1: Rewrite the Integral
We can express the integral in a more manageable form:
\[ I = \int \frac{dx}{(x-8)^{\frac{6}{7}}(x+6)^{\frac{8}{7}}} \]

### Step 2: Use Substitution
Let's use the substitution:
\[ t = \frac{x + 6}{x - 8} \]
This implies:
\[ x + 6 = t(x - 8) \quad \Rightarrow \quad x + 6 = tx - 8t \]
Rearranging gives:
\[ x(1 - t) = -8t - 6 \quad \Rightarrow \quad x = \frac{-8t - 6}{1 - t} \]

### Step 3: Differentiate to Find \(dx\)
Now, we differentiate \(t\) with respect to \(x\):
\[ \frac{dt}{dx} = \frac{(1)(x - 8) - (x + 6)(1)}{(x - 8)^2} = \frac{x - 8 - x - 6}{(x - 8)^2} = \frac{-14}{(x - 8)^2} \]
Thus,
\[ dx = -\frac{(x - 8)^2}{14} dt \]

### Step 4: Substitute Back into the Integral
Substituting \(dx\) into the integral:
\[ I = \int \frac{-\frac{(x - 8)^2}{14} dt}{(x - 8)^{\frac{6}{7}}(x + 6)^{\frac{8}{7}}} \]
This simplifies to:
\[ I = -\frac{1}{14} \int \frac{(x - 8)^{2 - \frac{6}{7}}}{(x + 6)^{\frac{8}{7}}} dt \]
\[ = -\frac{1}{14} \int \frac{(x - 8)^{\frac{8}{7}}}{(x + 6)^{\frac{8}{7}}} dt \]

### Step 5: Integrate
Now we can integrate:
\[ I = -\frac{1}{14} \cdot \frac{(x - 8)^{\frac{8}{7}}}{(x + 6)^{\frac{8}{7}}} + C \]

### Step 6: Back Substitute for \(t\)
Substituting back for \(t\):
\[ t = \frac{x + 6}{x - 8} \]
Thus, we can express the integral in terms of \(x\):
\[ I = -\frac{1}{14} \left(\frac{x + 6}{x - 8}\right) + C \]

### Final Answer
The final expression for the integral is:
\[ I = \frac{1}{2} \left(\frac{x - 8}{x + 6}\right) + C \]