The value of `int_(1)^(3)(sqrt(1+(x-1)^(3))+(x^(2)-1)^(1/3)+1)dx` is __________.

Explanation

To solve the definite integral

\[ I = \int_{1}^{3} \left( \sqrt{1 + (x - 1)^3} + (x^2 - 1)^{1/3} + 1 \right) dx, \]

we can follow these steps:

### Step 1: Simplify the integrand
We can rewrite the integrand as:

\[ \sqrt{1 + (x - 1)^3} + (x^2 - 1)^{1/3} + 1. \]

### Step 2: Make a substitution
Let’s define a function \( f(x) = \sqrt{1 + (x - 1)^3} + (x^2 - 1)^{1/3} + 1 \). We will evaluate the integral directly.

### Step 3: Evaluate the integral
We will compute the integral \( I \) directly:

\[ I = \int_{1}^{3} f(x) \, dx. \]

### Step 4: Calculate the integral
We can calculate the integral by evaluating the function at the limits:

1. **Evaluate \( f(1) \)**:
\[ f(1) = \sqrt{1 + (1 - 1)^3} + (1^2 - 1)^{1/3} + 1 = \sqrt{1} + 0 + 1 = 2. \]

2. **Evaluate \( f(3) \)**:
\[ f(3) = \sqrt{1 + (3 - 1)^3} + (3^2 - 1)^{1/3} + 1 = \sqrt{1 + 8} + (9 - 1)^{1/3} + 1 = \sqrt{9} + 2 + 1 = 3 + 2 + 1 = 6. \]

### Step 5: Apply the Fundamental Theorem of Calculus
Now, we can apply the Fundamental Theorem of Calculus. The integral can be computed as:

\[ I = \int_{1}^{3} f(x) \, dx = \left[ x \cdot f(x) \right]_{1}^{3} - \int_{1}^{3} f'(x) \, dx. \]

However, we can also directly calculate:

\[ I = f(3) - f(1) = 6 - 2 = 4. \]

### Step 6: Final Calculation
Thus, the value of the integral is:

\[ I = 4. \]

### Final Answer:
The value of \( \int_{1}^{3} \left( \sqrt{1 + (x - 1)^3} + (x^2 - 1)^{1/3} + 1 \right) dx \) is \( \boxed{4} \).