`"Let " k(x)=int((x^(2)+1)dx)/(root(3)(x^(3)+3x+6)) " and " k(-1)=(1)/(root(3)(2)). " Then the value of " k(-2) " is "-.`

Explanation

To solve the problem, we need to find the value of \( k(-2) \) given the function \( k(x) \) defined as:

\[ k(x) = \frac{\int (x^2 + 1) \, dx}{\sqrt[3]{x^3 + 3x + 6}} \]

We are also given that \( k(-1) = \frac{1}{\sqrt[3]{2}} \).

### Step 1: Find the integral \( \int (x^2 + 1) \, dx \)

The integral can be calculated as follows:

\[ \int (x^2 + 1) \, dx = \frac{x^3}{3} + x + C \]

### Step 2: Substitute the integral into \( k(x) \)

Now we can substitute this result back into the expression for \( k(x) \):

\[ k(x) = \frac{\frac{x^3}{3} + x + C}{\sqrt[3]{x^3 + 3x + 6}} \]

### Step 3: Simplify \( k(x) \)

We can express \( k(x) \) as:

\[ k(x) = \frac{x^3 + 3x + 6}{3\sqrt[3]{x^3 + 3x + 6}} + \frac{C}{\sqrt[3]{x^3 + 3x + 6}} \]

### Step 4: Use the given value \( k(-1) \)

We know that \( k(-1) = \frac{1}{\sqrt[3]{2}} \). Let's calculate \( k(-1) \):

1. Calculate \( x^3 + 3x + 6 \) at \( x = -1 \):

\[ (-1)^3 + 3(-1) + 6 = -1 - 3 + 6 = 2 \]

2. Substitute into \( k(-1) \):

\[ k(-1) = \frac{\frac{(-1)^3}{3} + (-1) + C}{\sqrt[3]{2}} = \frac{-\frac{1}{3} - 1 + C}{\sqrt[3]{2}} = \frac{C - \frac{4}{3}}{\sqrt[3]{2}} \]

Setting this equal to \( \frac{1}{\sqrt[3]{2}} \):

\[ \frac{C - \frac{4}{3}}{\sqrt[3]{2}} = \frac{1}{\sqrt[3]{2}} \]

### Step 5: Solve for \( C \)

Multiplying both sides by \( \sqrt[3]{2} \):

\[ C - \frac{4}{3} = 1 \]

Thus,

\[ C = 1 + \frac{4}{3} = \frac{7}{3} \]

### Step 6: Substitute \( C \) back into \( k(x) \)

Now we can write \( k(x) \):

\[ k(x) = \frac{\frac{x^3}{3} + x + \frac{7}{3}}{\sqrt[3]{x^3 + 3x + 6}} \]

### Step 7: Calculate \( k(-2) \)

1. Calculate \( x^3 + 3x + 6 \) at \( x = -2 \):

\[ (-2)^3 + 3(-2) + 6 = -8 - 6 + 6 = -8 \]

2. Substitute into \( k(-2) \):

\[ k(-2) = \frac{\frac{(-2)^3}{3} + (-2) + \frac{7}{3}}{\sqrt[3]{-8}} = \frac{-\frac{8}{3} - 2 + \frac{7}{3}}{-2} \]

3. Simplify the numerator:

\[ -\frac{8}{3} - \frac{6}{3} + \frac{7}{3} = -\frac{8 + 6 - 7}{3} = -\frac{7}{3} \]

Thus,

\[ k(-2) = \frac{-\frac{7}{3}}{-2} = \frac{7}{6} \]

### Final Result

The value of \( k(-2) \) is:

\[ \boxed{\frac{7}{6}} \]