`int (x^(2))/(1-2x^(3))dx`

Answer

To solve the integral \( \int \frac{x^2}{1 - 2x^3} \, dx \), we will follow these steps:

### Step 1: Substitution
Let \( t = 1 - 2x^3 \). Then, we need to find \( dt \):
\[ dt = -6x^2 \, dx \quad \Rightarrow \quad dx = \frac{dt}{-6x^2} \]

### Step 2: Express \( x^2 \, dx \) in terms of \( dt \)
From the equation \( dt = -6x^2 \, dx \), we can express \( x^2 \, dx \):
\[ x^2 \, dx = -\frac{1}{6} \, dt \]

### Step 3: Substitute in the integral
Now, substitute \( t \) and \( x^2 \, dx \) into the integral:
\[ \int \frac{x^2}{1 - 2x^3} \, dx = \int \frac{x^2}{t} \left(-\frac{1}{6} dt\right) = -\frac{1}{6} \int \frac{dt}{t} \]

### Step 4: Integrate
The integral \( \int \frac{dt}{t} \) is a standard integral:
\[ -\frac{1}{6} \int \frac{dt}{t} = -\frac{1}{6} \log |t| + C \]

### Step 5: Substitute back for \( t \)
Now, substitute back \( t = 1 - 2x^3 \):
\[ -\frac{1}{6} \log |1 - 2x^3| + C \]

### Final Answer
Thus, the final result for the integral is:
\[ \int \frac{x^2}{1 - 2x^3} \, dx = -\frac{1}{6} \log |1 - 2x^3| + C \]

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