`int (log(x + 1) - log x)/(x(x + 1)) dx` equals

Options

A. `-log ((x + 1)/(x)) + c`
B. `- log[log ((x + 1)/(x))] + c`
C. `- 1/2[log ((x +1)/(x))]^2 + c`
D. `c - 1/2 [log (x + 1)^2 - (log x)^2]`

Answer

To solve the integral \( \int \frac{\log(x + 1) - \log x}{x(x + 1)} \, dx \), we can follow these steps:

### Step 1: Simplify the integrand using logarithmic properties
We can use the property of logarithms that states \( \log a - \log b = \log \left( \frac{a}{b} \right) \). Thus, we can rewrite the integrand:

\[ \log(x + 1) - \log x = \log \left( \frac{x + 1}{x} \right) = \log \left( 1 + \frac{1}{x} \right) \]

Now, the integral becomes:

\[ \int \frac{\log \left( 1 + \frac{1}{x} \right)}{x(x + 1)} \, dx \]

### Step 2: Change of variables
Let \( t = \log \left( 1 + \frac{1}{x} \right) \). To find \( dt \), we differentiate:

\[ \frac{d}{dx} \left( \log \left( 1 + \frac{1}{x} \right) \right) = \frac{1}{1 + \frac{1}{x}} \cdot \left(-\frac{1}{x^2}\right) = -\frac{1}{x^2 + x} \]

Thus, we have:

\[ dt = -\frac{1}{x^2 + x} \, dx \quad \Rightarrow \quad dx = -(x^2 + x) \, dt \]

### Step 3: Substitute in the integral
Substituting \( t \) and \( dx \) into the integral gives:

\[ \int \frac{t}{x(x + 1)} \cdot -(x^2 + x) \, dt \]

The \( x(x + 1) \) in the denominator cancels with the \( x^2 + x \) in the numerator, leading to:

\[ -\int t \, dt \]

### Step 4: Integrate
Now we can integrate:

\[ -\int t \, dt = -\frac{t^2}{2} + C \]

### Step 5: Substitute back for \( t \)
Recall that \( t = \log \left( 1 + \frac{1}{x} \right) \). Thus, substituting back gives:

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

### Final Result
The final answer for the integral is:

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