What is `int_(a)^(b) (logx)/(x) dx` equal to ?

Options

A. `(1//2)log (ab).log(b/a)`
B. `log b//loga`
C. `log (b//a)`
D. `(1//2) log [(a+b)//b]`

Explanation

To solve the integral \( I = \int_{a}^{b} \frac{\log x}{x} \, dx \), we can use a substitution method. Let's go through the steps systematically.

### Step 1: Substitution
Let \( t = \log x \). Then, we differentiate to find \( dx \):
\[ \frac{dt}{dx} = \frac{1}{x} \implies dx = x \, dt = e^t \, dt \]
Now, we also need to change the limits of integration. When \( x = a \), \( t = \log a \) and when \( x = b \), \( t = \log b \).

### Step 2: Change the Integral
Substituting \( x \) and \( dx \) into the integral, we have:
\[ I = \int_{\log a}^{\log b} \frac{t}{e^t} e^t \, dt = \int_{\log a}^{\log b} t \, dt \]

### Step 3: Integrate
Now, we can integrate \( t \):
\[ \int t \, dt = \frac{t^2}{2} \]
Thus, we evaluate the definite integral:
\[ I = \left[ \frac{t^2}{2} \right]_{\log a}^{\log b} = \frac{(\log b)^2}{2} - \frac{(\log a)^2}{2} \]
This simplifies to:
\[ I = \frac{1}{2} \left( (\log b)^2 - (\log a)^2 \right) \]

### Step 4: Factor the Result
Using the difference of squares, we can factor the result:
\[ (\log b)^2 - (\log a)^2 = (\log b - \log a)(\log b + \log a) \]
Thus, we can write:
\[ I = \frac{1}{2} (\log b - \log a)(\log b + \log a) \]

### Step 5: Final Simplification
Using the property of logarithms \( \log b - \log a = \log \frac{b}{a} \) and \( \log b + \log a = \log(ab) \), we get:
\[ I = \frac{1}{2} \log \frac{b}{a} \cdot \log(ab) \]

### Final Answer
Therefore, the value of the integral is:
\[ \int_{a}^{b} \frac{\log x}{x} \, dx = \frac{1}{2} \log \frac{b}{a} \cdot \log(ab) \]