`int(log(x//e))/((logx)^(2))dx=`

Options

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

Answer

To solve the integral \( \int \frac{\log\left(\frac{x}{e}\right)}{(\log x)^2} \, dx \), we can follow these steps:

### Step 1: Simplify the logarithmic expression
Using the property of logarithms, we can rewrite \( \log\left(\frac{x}{e}\right) \) as:
\[ \log\left(\frac{x}{e}\right) = \log x - \log e \]
Since \( \log e = 1 \), we have:
\[ \log\left(\frac{x}{e}\right) = \log x - 1 \]
Thus, we can rewrite the integral as:
\[ \int \frac{\log x - 1}{(\log x)^2} \, dx \]

### Step 2: Split the integral
We can split the integral into two parts:
\[ \int \frac{\log x}{(\log x)^2} \, dx - \int \frac{1}{(\log x)^2} \, dx \]
This simplifies to:
\[ \int \frac{1}{\log x} \, dx - \int \frac{1}{(\log x)^2} \, dx \]

### Step 3: Substitution for the first integral
Let \( u = \log x \). Then, \( du = \frac{1}{x} \, dx \) or \( dx = e^u \, du \). The first integral becomes:
\[ \int \frac{1}{u} e^u \, du \]

### Step 4: Integration by parts for the first integral
Using integration by parts, let:
- \( v = e^u \) and \( dv = e^u \, du \)
- \( w = \frac{1}{u} \) and \( dw = -\frac{1}{u^2} \, du \)

Then, the integration by parts formula \( \int v \, dw = vw - \int w \, dv \) gives us:
\[ \int \frac{1}{u} e^u \, du = \frac{e^u}{u} - \int e^u \left(-\frac{1}{u^2}\right) \, du \]
This simplifies to:
\[ \frac{e^u}{u} + \int \frac{e^u}{u^2} \, du \]

### Step 5: Substitute back to \( x \)
Now, substituting back \( u = \log x \):
\[ \int \frac{1}{\log x} e^{\log x} \, dx = \frac{x}{\log x} + \int \frac{x}{(\log x)^2} \, dx \]

### Step 6: Combine the results
Now, we have:
\[ \int \frac{\log x - 1}{(\log x)^2} \, dx = \left( \frac{x}{\log x} + \int \frac{x}{(\log x)^2} \, dx \right) - \int \frac{1}{(\log x)^2} \, dx \]

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