If `y = int_(1)^(x) xsqrt(lnt)dt`
then find the value of `(d^(2)y)/(dx^(2))` at `x = e`

Answer

To solve the problem, we need to find the second derivative of the function \( y = \int_{1}^{x} x \sqrt{\ln t} \, dt \) at \( x = e \).

### Step-by-Step Solution:

1. **Define the function:**
\[ y = \int_{1}^{x} x \sqrt{\ln t} \, dt \]

2. **Differentiate \( y \) with respect to \( x \):**
We will use the product rule and Leibniz's rule for differentiation under the integral sign. Let \( u = x \) and \( v = \int_{1}^{x} \sqrt{\ln t} \, dt \).
\[ \frac{dy}{dx} = \frac{d}{dx}(x) \cdot \int_{1}^{x} \sqrt{\ln t} \, dt + x \cdot \frac{d}{dx}\left(\int_{1}^{x} \sqrt{\ln t} \, dt\right) \]
Using Leibniz's rule:
\[ \frac{d}{dx}\left(\int_{1}^{x} \sqrt{\ln t} \, dt\right) = \sqrt{\ln x} \]
Thus, we have:
\[ \frac{dy}{dx} = \int_{1}^{x} \sqrt{\ln t} \, dt + x \cdot \sqrt{\ln x} \]

3. **Differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \):**
Now we differentiate \( \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\int_{1}^{x} \sqrt{\ln t} \, dt\right) + \frac{d}{dx}(x \cdot \sqrt{\ln x}) \]
The first term is:
\[ \frac{d}{dx}\left(\int_{1}^{x} \sqrt{\ln t} \, dt\right) = \sqrt{\ln x} \]
For the second term, we apply the product rule:
\[ \frac{d}{dx}(x \cdot \sqrt{\ln x}) = \sqrt{\ln x} + x \cdot \frac{1}{2\sqrt{\ln x}} \cdot \frac{1}{x} = \sqrt{\ln x} + \frac{1}{2\sqrt{\ln x}} \]
Therefore, combining both parts:
\[ \frac{d^2y}{dx^2} = \sqrt{\ln x} + \sqrt{\ln x} + \frac{1}{2\sqrt{\ln x}} = 2\sqrt{\ln x} + \frac{1}{2\sqrt{\ln x}} \]

4. **Evaluate \( \frac{d^2y}{dx^2} \) at \( x = e \):**
We know that \( \ln e = 1 \):
\[ \frac{d^2y}{dx^2}\bigg|_{x=e} = 2\sqrt{\ln e} + \frac{1}{2\sqrt{\ln e}} = 2\sqrt{1} + \frac{1}{2\sqrt{1}} = 2 + \frac{1}{2} = 2 + 0.5 = \frac{5}{2} \]

### Final Answer:
\[ \frac{d^2y}{dx^2}\bigg|_{x=e} = \frac{5}{2} \]