Let `f(x)=int_(0)^(x)(dt)/(sqrt(1+t^(3)))` and `g(x)` be the inverse of `f(x)`. Then the value of `4 (g''(x))/(g(x)^(2))` is________.

Explanation

To solve the problem, we need to find the value of \( 4 \frac{g''(x)}{(g(x))^2} \) where \( g(x) \) is the inverse of the function \( f(x) = \int_0^x \frac{dt}{\sqrt{1 + t^3}} \).

### Step-by-step Solution:

1. **Understanding the Functions**:
We have \( f(x) = \int_0^x \frac{dt}{\sqrt{1 + t^3}} \) and \( g(x) \) is the inverse of \( f(x) \). This means that \( f(g(x)) = x \).

2. **Differentiating Both Sides**:
Differentiate \( f(g(x)) = x \) with respect to \( x \):
\[ f'(g(x)) \cdot g'(x) = 1 \]
From this, we can express \( g'(x) \):
\[ g'(x) = \frac{1}{f'(g(x))} \]

3. **Finding \( f'(x) \)**:
By applying the Leibniz rule, we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx} \left( \int_0^x \frac{dt}{\sqrt{1 + t^3}} \right) = \frac{1}{\sqrt{1 + x^3}} \]
Therefore, substituting \( g(x) \) for \( x \):
\[ f'(g(x)) = \frac{1}{\sqrt{1 + (g(x))^3}} \]

4. **Substituting Back**:
Now substituting back into the equation for \( g'(x) \):
\[ g'(x) = \sqrt{1 + (g(x))^3} \]

5. **Finding \( g''(x) \)**:
Differentiate \( g'(x) \) to find \( g''(x) \):
\[ g''(x) = \frac{d}{dx} \left( \sqrt{1 + (g(x))^3} \right) \]
Using the chain rule:
\[ g''(x) = \frac{1}{2\sqrt{1 + (g(x))^3}} \cdot 3(g(x))^2 g'(x) \]
Substitute \( g'(x) \):
\[ g''(x) = \frac{3(g(x))^2 g'(x)}{2\sqrt{1 + (g(x))^3}} = \frac{3(g(x))^2 \sqrt{1 + (g(x))^3}}{2\sqrt{1 + (g(x))^3}} \]
Simplifying gives:
\[ g''(x) = \frac{3}{2} g(x)^2 \]

6. **Finding \( \frac{g''(x)}{(g(x))^2} \)**:
Now we can find:
\[ \frac{g''(x)}{(g(x))^2} = \frac{\frac{3}{2} g(x)^2}{(g(x))^2} = \frac{3}{2} \]

7. **Final Calculation**:
Finally, we need to find \( 4 \frac{g''(x)}{(g(x))^2} \):
\[ 4 \frac{g''(x)}{(g(x))^2} = 4 \cdot \frac{3}{2} = 6 \]

Thus, the value of \( 4 \frac{g''(x)}{(g(x))^2} \) is **6**.