What is `int_a^b (log x)/x dx` equal to

Options

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

Answer

To solve the integral \( I = \int_a^b \frac{\log x}{x} \, dx \), we can use a substitution method. Here’s a step-by-step solution:

### Step 1: Set up the integral
We start with the integral:
\[ I = \int_a^b \frac{\log x}{x} \, dx \]

### Step 2: Use substitution
Let’s use the substitution:
\[ t = \log x \]
Then, the differential \( dx \) can be expressed in terms of \( dt \):
\[ dx = e^t \, dt \]
Also, since \( x = e^t \), we have:
\[ \frac{1}{x} = \frac{1}{e^t} \]
Thus, the integral becomes:
\[ I = \int \frac{t}{e^t} e^t \, dt = \int t \, dt \]

### Step 3: Change the limits of integration
Next, we need to change the limits of integration. When \( x = a \):
\[ t = \log a \]
When \( x = b \):
\[ t = \log b \]
So the new limits are from \( \log a \) to \( \log b \).

### Step 4: Evaluate the integral
Now we evaluate the integral:
\[ I = \int_{\log a}^{\log b} t \, dt \]
The integral of \( t \) is:
\[ \frac{t^2}{2} \]
Thus, we have:
\[ I = \left[ \frac{t^2}{2} \right]_{\log a}^{\log b} \]
Calculating this gives:
\[ I = \frac{(\log b)^2}{2} - \frac{(\log a)^2}{2} \]

### Step 5: Simplify the result
This can be simplified as:
\[ I = \frac{1}{2} \left( (\log b)^2 - (\log a)^2 \right) \]
Using the identity \( a^2 - b^2 = (a-b)(a+b) \), we can write:
\[ I = \frac{1}{2} \left( \log b - \log a \right) \left( \log b + \log a \right) \]
This can also be expressed using properties of logarithms:
\[ I = \frac{1}{2} \log \left( \frac{b}{a} \right) \log(ab) \]

### Final Result
Thus, the final result for the integral is:
\[ I = \frac{1}{2} \left( \log b \right)^2 - \frac{1}{2} \left( \log a \right)^2 \]