Answer is: option1

Solution:

We are given:

\( F(x) = \int_0^{x^2} \frac{\sqrt{t^2 + 3}}{2t} \, dt \)

We are to find:

\( F''(1) \)

Let:

1. \( f(t) = \frac{\sqrt{t^2 + 3}}{2t} \)
2. \( u(x) = x^2 \)

Then:

\( F'(x) = f(u(x)) \cdot u'(x) = \frac{\sqrt{x^4 + 3}}{2x^2} \cdot 2x = \frac{x \sqrt{x^4 + 3}}{x^2} = \frac{\sqrt{x^4 + 3}}{x} \)

We use the quotient rule:

Let:

1. \( u = \sqrt{x^4 + 3} \)
2. \( v = x \)

Then:

\( F''(x) = \frac{v \cdot u' - u \cdot v'}{v^2} \)

First compute:

1. \( u = (x^4 + 3)^{1/2} \Rightarrow u' = \frac{1}{2}(x^4 + 3)^{-1/2} \cdot 4x^3 = \frac{2x^3}{\sqrt{x^4 + 3}} \)
2. \( v = x \Rightarrow v' = 1 \)

Now plug in:

\( F''(x) = \frac{x \cdot \frac{2x^3}{\sqrt{x^4 + 3}} - \sqrt{x^4 + 3} \cdot 1}{x^2} = \frac{2x^4 - (x^4 + 3)}{x^2 \cdot \sqrt{x^4 + 3}} \)

\( F''(1) = \frac{2(1)^4 - (1^4 + 3)}{1^2 \cdot \sqrt{1^4 + 3}} = \frac{2 - 4}{\sqrt{4}} = \frac{-2}{2} = -1 \)

Hence option A