Let A `= int_(1)^(e^(2))(lnx)/(sqrt(x))dx`, then

Options

A. `A gt 2 (e-1/e)`
B. `A lt (e-1)(2+1/(sqrt(e)))`
C. `A gt (e-1) (2+(1)/(sqrt(e )))`
D. `A lt (e^(2)-1)'2/e`

Answer

To solve the integral \( A = \int_{1}^{e^2} \frac{\ln x}{\sqrt{x}} \, dx \), we will use a substitution method.

### Step-by-step Solution:

1. **Substitution**: Let \( t = \sqrt{x} \). Then, \( x = t^2 \) and \( dx = 2t \, dt \).
- When \( x = 1 \), \( t = \sqrt{1} = 1 \).
- When \( x = e^2 \), \( t = \sqrt{e^2} = e \).

Thus, the limits of integration change from \( x = 1 \) to \( x = e^2 \) into \( t = 1 \) to \( t = e \).

2. **Rewrite the Integral**: Substitute \( x \) and \( dx \) in the integral:
\[ A = \int_{1}^{e} \frac{\ln(t^2)}{t} \cdot 2t \, dt = 2 \int_{1}^{e} \ln(t^2) \, dt \]

3. **Simplify the Logarithm**: Use the property of logarithms, \( \ln(t^2) = 2 \ln(t) \):
\[ A = 2 \int_{1}^{e} 2 \ln(t) \, dt = 4 \int_{1}^{e} \ln(t) \, dt \]

4. **Integration by Parts**: Let \( u = \ln(t) \) and \( dv = dt \).
- Then, \( du = \frac{1}{t} dt \) and \( v = t \).
- Using integration by parts, \( \int u \, dv = uv - \int v \, du \):
\[ \int \ln(t) \, dt = t \ln(t) - \int t \cdot \frac{1}{t} \, dt = t \ln(t) - t \]

5. **Evaluate the Integral**:
\[ \int_{1}^{e} \ln(t) \, dt = \left[ t \ln(t) - t \right]_{1}^{e} \]
- Evaluating at the limits:
\[ = \left[ e \ln(e) - e \right] - \left[ 1 \ln(1) - 1 \right] = \left[ e \cdot 1 - e \right] - \left[ 0 - 1 \right] = 0 + 1 = 1 \]

6. **Final Calculation**:
\[ A = 4 \cdot 1 = 4 \]

### Conclusion:
Thus, the value of the integral \( A \) is \( 4 \).