`"If " f(x)=int(dx)/(x^(1//3)+2) " and "f(0)=12log_(e)2, " then the value of " f(-1) " is"-.`

Explanation

To solve the problem, we need to find the value of \( f(-1) \) given that

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

and \( f(0) = 12 \ln 2 \).

### Step 1: Change of Variable
Let's start by making a substitution to simplify the integral. Let

\[ t^3 = x \implies dx = 3t^2 dt. \]

### Step 2: Rewrite the Integral
Substituting \( x \) in the integral, we have:

\[ f(x) = \int \frac{3t^2 dt}{t + 2}. \]

### Step 3: Simplify the Integral
Now, we can separate the integral:

\[ f(x) = 3 \int \frac{t^2}{t + 2} dt. \]

To simplify \( \frac{t^2}{t + 2} \), we can perform polynomial long division:

\[ \frac{t^2}{t + 2} = t - 2 + \frac{4}{t + 2}. \]

### Step 4: Integrate Each Term
Now we can integrate:

\[ f(x) = 3 \left( \int (t - 2) dt + \int \frac{4}{t + 2} dt \right). \]

Calculating each integral:

1. \( \int (t - 2) dt = \frac{t^2}{2} - 2t \)
2. \( \int \frac{4}{t + 2} dt = 4 \ln |t + 2| \)

Thus,

\[ f(x) = 3 \left( \frac{t^2}{2} - 2t + 4 \ln |t + 2| \right) + C. \]

### Step 5: Substitute Back for \( t \)
Substituting back \( t = x^{1/3} \):

\[ f(x) = 3 \left( \frac{x^{2/3}}{2} - 2x^{1/3} + 4 \ln |x^{1/3} + 2| \right) + C. \]

### Step 6: Use the Given Condition
We know \( f(0) = 12 \ln 2 \). Let's find \( f(0) \):

\[ f(0) = 3 \left( 0 - 0 + 4 \ln |2| \right) + C = 12 \ln 2. \]

This implies:

\[ 12 \ln 2 + C = 12 \ln 2 \implies C = 0. \]

### Step 7: Final Expression for \( f(x) \)
Thus, we have:

\[ f(x) = \frac{3}{2} x^{2/3} - 6 x^{1/3} + 12 \ln |x^{1/3} + 2|. \]

### Step 8: Calculate \( f(-1) \)
Now, we need to find \( f(-1) \):

\[ f(-1) = \frac{3}{2} (-1)^{2/3} - 6 (-1)^{1/3} + 12 \ln |-1^{1/3} + 2|. \]

Calculating each term:

1. \( (-1)^{2/3} = 1 \)
2. \( (-1)^{1/3} = -1 \)
3. \( |-1^{1/3} + 2| = | -1 + 2 | = 1 \)

So,

\[ f(-1) = \frac{3}{2} \cdot 1 - 6 \cdot (-1) + 12 \ln 1. \]

Since \( \ln 1 = 0 \):

\[ f(-1) = \frac{3}{2} + 6 = \frac{3}{2} + \frac{12}{2} = \frac{15}{2}. \]

### Final Answer
Thus, the value of \( f(-1) \) is

\[ \boxed{\frac{15}{2}}. \]