`int log x*[log(ex)]^(-2)dx=` . . .

Options

A. `(x)/(1+logx)+c`
B. `x(1-logx)+c`
C. `x(1+logx)+c`
D. `(x)/(1-logx)+c`

Explanation

To solve the integral \( \int \log x \cdot [\log(e^x)]^{-2} \, dx \), we will follow these steps:

### Step 1: Simplify the integrand
We start with the integral:
\[ \int \log x \cdot [\log(e^x)]^{-2} \, dx \]
We know that \( \log(e^x) = x \cdot \log e = x \cdot 1 = x \). Therefore, we can rewrite the integrand:
\[ \int \log x \cdot \frac{1}{x^2} \, dx \]
This simplifies to:
\[ \int \frac{\log x}{x^2} \, dx \]

### Step 2: Use substitution
Let \( t = \log x \). Then, we have:
\[ x = e^t \quad \text{and} \quad dx = e^t \, dt \]
Substituting these into the integral gives:
\[ \int \frac{t}{(e^t)^2} \cdot e^t \, dt = \int \frac{t}{e^t} \, dt \]

### Step 3: Integration by parts
We will use integration by parts, where we let:
- \( u = t \) (thus \( du = dt \))
- \( dv = e^{-t} dt \) (thus \( v = -e^{-t} \))

Applying integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
Substituting in our values:
\[ \int \frac{t}{e^t} \, dt = -t e^{-t} - \int -e^{-t} \, dt \]
This simplifies to:
\[ -t e^{-t} + e^{-t} + C \]
where \( C \) is the constant of integration.

### Step 4: Substitute back
Now we substitute back \( t = \log x \):
\[ -\log x \cdot \frac{1}{x} + \frac{1}{x} + C \]
This can be rewritten as:
\[ \frac{1 - \log x}{x} + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \log x \cdot [\log(e^x)]^{-2} \, dx = \frac{1 - \log x}{x} + C \]