Prove that the function L(x) defined on the interval `(0,infty)` by the integral `L(x) = int_(1)^(x) (dt)/(t)` possesses the following properties.
`L(x_(1)x_(2)) = L(x_(1)) + L(x_(2))`
`L((x_(1))/(x_(2))) = L(x_(1)) - L(x_(2))`

Answer

To prove the properties of the function \( L(x) \) defined by the integral

\[ L(x) = \int_{1}^{x} \frac{dt}{t} \]

we will demonstrate the two properties step by step.

### Property 1: \( L(x_1 x_2) = L(x_1) + L(x_2) \)

1. **Start with the definition of \( L(x) \)**:
\[ L(x_1 x_2) = \int_{1}^{x_1 x_2} \frac{dt}{t} \]

2. **Use the property of definite integrals**:
We can split the integral from \( 1 \) to \( x_1 x_2 \) into two parts:
\[ L(x_1 x_2) = \int_{1}^{x_1} \frac{dt}{t} + \int_{x_1}^{x_1 x_2} \frac{dt}{t} \]

3. **Change of variable in the second integral**:
Let \( t = x_1 u \) where \( u \) goes from \( 1 \) to \( x_2 \) when \( t \) goes from \( x_1 \) to \( x_1 x_2 \). The differential \( dt = x_1 du \):
\[ \int_{x_1}^{x_1 x_2} \frac{dt}{t} = \int_{1}^{x_2} \frac{x_1 du}{x_1 u} = \int_{1}^{x_2} \frac{du}{u} \]

4. **Combine the results**:
Thus, we have:
\[ L(x_1 x_2) = \int_{1}^{x_1} \frac{dt}{t} + \int_{1}^{x_2} \frac{du}{u} = L(x_1) + L(x_2) \]

### Property 2: \( L\left(\frac{x_1}{x_2}\right) = L(x_1) - L(x_2) \)

1. **Start with the definition of \( L(x) \)**:
\[ L\left(\frac{x_1}{x_2}\right) = \int_{1}^{\frac{x_1}{x_2}} \frac{dt}{t} \]

2. **Change of variable**:
Let \( t = \frac{x_1}{u} \) where \( u \) goes from \( x_2 \) to \( x_1 \) when \( t \) goes from \( 1 \) to \( \frac{x_1}{x_2} \). The differential \( dt = -\frac{x_1}{u^2} du \):
\[ L\left(\frac{x_1}{x_2}\right) = \int_{x_2}^{x_1} \frac{-\frac{x_1}{u^2} du}{\frac{x_1}{u}} = \int_{x_2}^{x_1} \frac{-du}{u} \]

3. **Evaluate the integral**:
This can be rewritten as:
\[ L\left(\frac{x_1}{x_2}\right) = -\left(\int_{1}^{x_1} \frac{du}{u} - \int_{1}^{x_2} \frac{du}{u}\right) = L(x_1) - L(x_2) \]

### Conclusion

Thus, we have proven both properties:
1. \( L(x_1 x_2) = L(x_1) + L(x_2) \)
2. \( L\left(\frac{x_1}{x_2}\right) = L(x_1) - L(x_2) \)