`int(x^(2))/(1+x^(6)) dx`

Answer

To solve the integral \( \int \frac{x^2}{1 + x^6} \, dx \), we will use substitution. Here are the steps:

### Step-by-Step Solution

1. **Identify the Integral**:
\[ I = \int \frac{x^2}{1 + x^6} \, dx \]

2. **Substitution**:
Let \( t = x^3 \). Then, differentiate both sides:
\[ dt = 3x^2 \, dx \quad \Rightarrow \quad dx = \frac{dt}{3x^2} \]
From the substitution \( t = x^3 \), we can express \( x^2 \) in terms of \( t \):
\[ x^2 = t^{2/3} \]

3. **Rewrite the Integral**:
Substitute \( x^2 \) and \( dx \) into the integral:
\[ I = \int \frac{t^{2/3}}{1 + (t)^2} \cdot \frac{dt}{3t^{2/3}} = \int \frac{1}{3(1 + t^2)} \, dt \]

4. **Factor out the Constant**:
\[ I = \frac{1}{3} \int \frac{1}{1 + t^2} \, dt \]

5. **Integrate**:
The integral of \( \frac{1}{1 + t^2} \) is \( \tan^{-1}(t) \):
\[ I = \frac{1}{3} \tan^{-1}(t) + C \]

6. **Back Substitute**:
Replace \( t \) back with \( x^3 \):
\[ I = \frac{1}{3} \tan^{-1}(x^3) + C \]

### Final Answer:
\[ \int \frac{x^2}{1 + x^6} \, dx = \frac{1}{3} \tan^{-1}(x^3) + C \]