`int \ {f(x)*g^(prime)(x)-f^(prime)(x)g(x))/(f(x)*g(x)){logg(x)-logf(x)} \ dx`

Options

A. `log"" ((g(x ))/(f (x)))+C`
B. `1/2 ((g (x ))/(f (x)))^(2)+C`
C. `1/2 (log ((g (x ))/(f (x))))^(2)+C`
D. `log (((g (x))/(f (x)))^(2))+C`

Answer

To solve the integral

\[ \int \frac{f(x) g'(x) - f'(x) g(x)}{f(x) g(x) \left( \log g(x) - \log f(x) \right)} \, dx, \]

we will follow these steps:

### Step 1: Simplify the logarithmic expression
Using the property of logarithms, we can rewrite the expression \(\log g(x) - \log f(x)\) as:

\[ \log \left( \frac{g(x)}{f(x)} \right). \]

Thus, the integral becomes:

\[ \int \frac{f(x) g'(x) - f'(x) g(x)}{f(x) g(x) \log \left( \frac{g(x)}{f(x)} \right)} \, dx. \]

**Hint:** Remember that \(\log a - \log b = \log \left( \frac{a}{b} \right)\).

### Step 2: Substitute \(t\)
Let

\[ t = \log \left( \frac{g(x)}{f(x)} \right). \]

Now, differentiate \(t\):

\[ dt = \frac{1}{g(x)} g'(x) - \frac{1}{f(x)} f'(x) \, dx. \]

Rearranging gives:

\[ dt = \left( \frac{g'(x)}{g(x)} - \frac{f'(x)}{f(x)} \right) \, dx. \]

### Step 3: Express \(f(x) g'(x) - f'(x) g(x)\) in terms of \(dt\)
From the expression for \(dt\), we can rewrite \(f(x) g'(x) - f'(x) g(x)\) in terms of \(dt\):

\[ f(x) g'(x) - f'(x) g(x) = f(x) g(x) \left( \frac{g'(x)}{g(x)} - \frac{f'(x)}{f(x)} \right) = f(x) g(x) dt. \]

### Step 4: Substitute back into the integral
Now substitute this back into the integral:

\[ \int \frac{f(x) g(x) dt}{f(x) g(x) t} = \int \frac{dt}{t}. \]

### Step 5: Integrate
The integral of \(\frac{1}{t}\) is:

\[ \int \frac{dt}{t} = \log |t| + C. \]

### Step 6: Substitute \(t\) back
Now substitute \(t\) back:

\[ \log |t| + C = \log \left| \log \left( \frac{g(x)}{f(x)} \right) \right| + C. \]

### Final Result
Thus, the final result of the integral is:

\[ \frac{1}{2} \left( \log \left( \frac{g(x)}{f(x)} \right) \right)^2 + C. \]

### Conclusion
The correct option is:

\[ \frac{1}{2} \left( \log \left( \frac{g(x)}{f(x)} \right) \right)^2 + C. \]