`int_(e)^(e^(2)) (1/logx-1/((logx)^(2)))dx=`

Options

A. `(e(e-1))/2`
B. `(-e(e-1))/2`
C. `(e(e-2))/2`
D. `(-e(e-2))/2`

Answer

To solve the integral

\[ \int_{e}^{e^2} \left( \frac{1}{\log x} - \frac{1}{(\log x)^2} \right) dx, \]

we can break it down into two separate integrals:

\[ \int_{e}^{e^2} \frac{1}{\log x} \, dx - \int_{e}^{e^2} \frac{1}{(\log x)^2} \, dx. \]

### Step 1: Solve the first integral \(\int_{e}^{e^2} \frac{1}{\log x} \, dx\)

To solve this integral, we can use integration by parts. Let:

- \( u = \log x \) \(\Rightarrow du = \frac{1}{x} dx\)
- \( dv = dx \) \(\Rightarrow v = x\)

Using integration by parts:

\[ \int u \, dv = uv - \int v \, du, \]

we have:

\[ \int \frac{1}{\log x} \, dx = x \log x - \int x \cdot \frac{1}{x \log x} \, dx = x \log x - \int \frac{1}{\log x} \, dx. \]

However, we need to evaluate it from \(e\) to \(e^2\). So, we directly compute:

\[ \int_{e}^{e^2} \frac{1}{\log x} \, dx = \left[ x \log x \right]_{e}^{e^2}. \]

Calculating the limits:

\[ = \left( e^2 \cdot \log(e^2) \right) - \left( e \cdot \log e \right) = e^2 \cdot 2 - e \cdot 1 = 2e^2 - e. \]

### Step 2: Solve the second integral \(\int_{e}^{e^2} \frac{1}{(\log x)^2} \, dx\)

For the second integral, we can again use integration by parts. Let:

- \( u = \frac{1}{\log x} \) \(\Rightarrow du = -\frac{1}{(\log x)^2} \cdot \frac{1}{x} dx\)
- \( dv = dx \) \(\Rightarrow v = x\)

Using integration by parts:

\[ \int u \, dv = uv - \int v \, du, \]

we have:

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

Evaluating from \(e\) to \(e^2\):

\[ \int_{e}^{e^2} \frac{1}{(\log x)^2} \, dx = \left[ \frac{x}{\log x} \right]_{e}^{e^2}. \]

Calculating the limits:

\[ = \left( \frac{e^2}{\log(e^2)} \right) - \left( \frac{e}{\log e} \right) = \frac{e^2}{2} - e. \]

### Step 3: Combine the results

Now we combine the results of both integrals:

\[ \int_{e}^{e^2} \left( \frac{1}{\log x} - \frac{1}{(\log x)^2} \right) dx = \left( 2e^2 - e \right) - \left( \frac{e^2}{2} - e \right). \]

Simplifying this:

\[ = 2e^2 - e - \frac{e^2}{2} + e = 2e^2 - \frac{e^2}{2} = \frac{4e^2}{2} - \frac{e^2}{2} = \frac{3e^2}{2}. \]

### Final Answer

Thus, the value of the integral is:

\[ \frac{3e^2}{2}. \]