To find the value of `int(1+logx)/(x)dx`, the proper substitution is

Options

A. `logx=t`
B. `1+logx=t`
C. `(1)/(x)=t`
D. `x=t`

Answer

To solve the integral \( \int \frac{1 + \log x}{x} \, dx \), we can use a substitution method. Here’s the step-by-step solution:

### Step 1: Identify the substitution
Let’s set:
\[ t = 1 + \log x \]

### Step 2: Differentiate the substitution
Now, we need to find \( dt \) in terms of \( dx \):
\[ \frac{dt}{dx} = \frac{d}{dx}(1 + \log x) = \frac{1}{x} \]
Thus, we can express \( dt \) as:
\[ dt = \frac{1}{x} \, dx \quad \Rightarrow \quad dx = x \, dt \]

### Step 3: Express \( x \) in terms of \( t \)
From our substitution \( t = 1 + \log x \), we can express \( \log x \) as:
\[ \log x = t - 1 \quad \Rightarrow \quad x = e^{t-1} \]

### Step 4: Substitute in the integral
Now, substituting \( t \) and \( dx \) into the integral:
\[ \int \frac{1 + \log x}{x} \, dx = \int \frac{t}{x} \cdot x \, dt = \int t \, dt \]

### Step 5: Integrate
Now we can integrate:
\[ \int t \, dt = \frac{t^2}{2} + C \]

### Step 6: Substitute back for \( t \)
Now, substituting back \( t = 1 + \log x \):
\[ \frac{(1 + \log x)^2}{2} + C \]

### Final Result
Thus, the value of the integral is:
\[ \int \frac{1 + \log x}{x} \, dx = \frac{(1 + \log x)^2}{2} + C \]

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