If `F(x)=int_(1)^(x) f(t) dt`,where `f(t)=int_(1)^(t^(2))(sqrt(1+u^(4)))/(u) du`, then the value of F''(2) equals to

Options

A. `(7)/(4sqrt17`
B. `(15)/(sqrt17)`
C. `sqrt(257)`
D. `(15sqrt17)/(68)`

Answer

To find the value of \( F''(2) \) where \( F(x) = \int_{1}^{x} f(t) \, dt \) and \( f(t) = \int_{1}^{t^2} \frac{\sqrt{1 + u^4}}{u} \, du \), we will follow these steps:

### Step 1: Differentiate \( F(x) \) to find \( F'(x) \)
Using the Fundamental Theorem of Calculus, we have:
\[ F'(x) = f(x) \]

### Step 2: Differentiate \( F'(x) \) to find \( F''(x) \)
Again applying the Fundamental Theorem of Calculus:
\[ F''(x) = f'(x) \]

### Step 3: Find \( f'(t) \)
To find \( f'(t) \), we need to differentiate \( f(t) \):
\[ f(t) = \int_{1}^{t^2} \frac{\sqrt{1 + u^4}}{u} \, du \]
Using the Leibniz rule for differentiation under the integral sign:
\[ f'(t) = \frac{d}{dt} \left( \int_{1}^{t^2} \frac{\sqrt{1 + u^4}}{u} \, du \right) = \frac{\sqrt{1 + (t^2)^4}}{t^2} \cdot \frac{d}{dt}(t^2) - 0 \]
This simplifies to:
\[ f'(t) = \frac{\sqrt{1 + t^8}}{t^2} \cdot 2t = \frac{2t \sqrt{1 + t^8}}{t^2} = \frac{2\sqrt{1 + t^8}}{t} \]

### Step 4: Evaluate \( F''(2) \)
Now we need to evaluate \( F''(2) \):
\[ F''(2) = f'(2) = \frac{2\sqrt{1 + 2^8}}{2} = \sqrt{1 + 256} = \sqrt{257} \]

### Final Answer
Thus, the value of \( F''(2) \) is:
\[ \sqrt{257} \]

---