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

Explanation

To solve the integral \( \int_{1}^{3} \left( \sqrt{1 + (x - 1)^3} + (x^2 - 1)^{1/3} + 1 \right) dx \), we will follow these steps:

### Step 1: Define the function
Let \( f(x) = \sqrt{1 + (x - 1)^3} + (x^2 - 1)^{1/3} + 1 \).

### Step 2: Find the inverse function
To find the inverse function, we note that if we let \( y = f(x) \), then we can express \( x \) in terms of \( y \). However, for this problem, we will not explicitly find the inverse function but will use the property of definite integrals.

### Step 3: Use symmetry in the integral
We can observe that \( f(x) \) and \( f(4 - x) \) might have a symmetric property that can simplify our calculations. We compute \( f(4 - x) \):

\[ f(4 - x) = \sqrt{1 + (4 - x - 1)^3} + ((4 - x)^2 - 1)^{1/3} + 1 \]
\[ = \sqrt{1 + (3 - x)^3} + ((16 - 8x + x^2) - 1)^{1/3} + 1 \]
\[ = \sqrt{1 + (3 - x)^3} + (15 - 8x + x^2)^{1/3} + 1 \]

### Step 4: Combine the integrals
Using the property of definite integrals, we can write:
\[ \int_{1}^{3} f(x) \, dx + \int_{1}^{3} f(4 - x) \, dx = \int_{1}^{3} (f(x) + f(4 - x)) \, dx \]

### Step 5: Evaluate the integral
Since \( f(x) + f(4 - x) \) simplifies to a constant, we can find the average value:
\[ \int_{1}^{3} f(x) \, dx = \int_{1}^{3} f(4 - x) \, dx \]
This means:
\[ 2 \int_{1}^{3} f(x) \, dx = \int_{1}^{3} (f(x) + f(4 - x)) \, dx \]

### Step 6: Calculate \( f(1) \) and \( f(3) \)
Now we calculate \( f(1) \) and \( f(3) \):
- For \( f(1) \):
\[ f(1) = \sqrt{1 + (1 - 1)^3} + (1^2 - 1)^{1/3} + 1 = \sqrt{1} + 0 + 1 = 2 \]
- For \( f(3) \):
\[ f(3) = \sqrt{1 + (3 - 1)^3} + (3^2 - 1)^{1/3} + 1 = \sqrt{1 + 8} + (9 - 1)^{1/3} + 1 = \sqrt{9} + 2 + 1 = 3 + 2 + 1 = 6 \]

### Step 7: Final calculation
Using the average value:
\[ \int_{1}^{3} f(x) \, dx = \frac{1}{2} \left( f(1) + f(3) \right) \cdot (3 - 1) \]
\[ = \frac{1}{2} \left( 2 + 6 \right) \cdot 2 = \frac{8}{2} \cdot 2 = 4 \cdot 2 = 8 \]

### Final Answer
Thus, the value of the integral is \( \boxed{8} \).