Let f be a function such that `f(xy) = f(x).f(y), AA y in R and R(1 + x) = 1 + x(1 + g(x))`. where `lim_(x rarr 0) g(x) = 0`. Find the value of `int_(1)^(2) (f(x))/(f'(x)).(1)/(1+x^(2))dx`

Answer

To solve the problem, we need to find the value of the integral

\[ \int_{1}^{2} \frac{f(x)}{f'(x)} \cdot \frac{1}{1+x^2} \, dx \]

given the properties of the function \( f \).

### Step 1: Determine the function \( f(x) \)

We know that \( f(xy) = f(x)f(y) \) for all \( y \in \mathbb{R} \). A common function that satisfies this property is the exponential function. Let's assume:

\[ f(x) = e^{g(x)} \]

where \( g(x) \) is some function. Then, we have:

\[ f(xy) = e^{g(xy)} = e^{g(x) + g(y)} = f(x)f(y) \]

This implies that \( g(xy) = g(x) + g(y) \). A well-known solution to this functional equation is \( g(x) = k \ln(x) \) for some constant \( k \). Therefore, we can express \( f(x) \) as:

\[ f(x) = e^{k \ln(x)} = x^k \]

### Step 2: Differentiate \( f(x) \)

Now, we differentiate \( f(x) \):

\[ f'(x) = kx^{k-1} \]

### Step 3: Substitute \( f(x) \) and \( f'(x) \) into the integral

Now we substitute \( f(x) \) and \( f'(x) \) into the integral:

\[ \int_{1}^{2} \frac{f(x)}{f'(x)} \cdot \frac{1}{1+x^2} \, dx = \int_{1}^{2} \frac{x^k}{kx^{k-1}} \cdot \frac{1}{1+x^2} \, dx \]

This simplifies to:

\[ \int_{1}^{2} \frac{x}{k} \cdot \frac{1}{1+x^2} \, dx = \frac{1}{k} \int_{1}^{2} \frac{x}{1+x^2} \, dx \]

### Step 4: Evaluate the integral

To evaluate the integral \( \int \frac{x}{1+x^2} \, dx \), we can use the substitution \( u = 1 + x^2 \), which gives \( du = 2x \, dx \) or \( dx = \frac{du}{2x} \).

Thus, we have:

\[ \int \frac{x}{1+x^2} \, dx = \frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln|u| + C = \frac{1}{2} \ln(1+x^2) + C \]

Now we evaluate the definite integral:

\[ \int_{1}^{2} \frac{x}{1+x^2} \, dx = \left[ \frac{1}{2} \ln(1+x^2) \right]_{1}^{2} = \frac{1}{2} \left( \ln(5) - \ln(2) \right) = \frac{1}{2} \ln\left(\frac{5}{2}\right) \]

### Step 5: Final result

Putting it all together, we have:

\[ \int_{1}^{2} \frac{f(x)}{f'(x)} \cdot \frac{1}{1+x^2} \, dx = \frac{1}{k} \cdot \frac{1}{2} \ln\left(\frac{5}{2}\right) \]

Assuming \( k = 1 \) for simplicity, the final answer is:

\[ \frac{1}{2} \ln\left(\frac{5}{2}\right) \]