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

Answer

To solve the integral \( I = \int \tan^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right) dx \), we can utilize a trigonometric identity.

### Step-by-step Solution:

1. **Recognize the Identity**:
We know that:
\[ \tan^{-1}(3x - x^3) = 3 \tan^{-1}(x) \]
Therefore, we can rewrite the integral as:
\[ I = \int 3 \tan^{-1}(x) \, dx \]

2. **Factor Out the Constant**:
We can factor out the constant 3 from the integral:
\[ I = 3 \int \tan^{-1}(x) \, dx \]

3. **Integration by Parts**:
We will use integration by parts, where we let:
- \( u = \tan^{-1}(x) \) (hence \( du = \frac{1}{1+x^2} \, dx \))
- \( dv = dx \) (hence \( v = x \))

The integration by parts formula is:
\[ \int u \, dv = uv - \int v \, du \]
Applying this, we have:
\[ \int \tan^{-1}(x) \, dx = x \tan^{-1}(x) - \int x \cdot \frac{1}{1+x^2} \, dx \]

4. **Simplify the Remaining Integral**:
The remaining integral can be simplified:
\[ \int x \cdot \frac{1}{1+x^2} \, dx = \int \frac{x}{1+x^2} \, dx \]
We can use the substitution \( t = 1 + x^2 \) (thus \( dt = 2x \, dx \) or \( dx = \frac{dt}{2x} \)):
\[ \int \frac{x}{1+x^2} \, dx = \frac{1}{2} \int \frac{1}{t} \, dt = \frac{1}{2} \ln |t| + C = \frac{1}{2} \ln(1+x^2) + C \]

5. **Combine the Results**:
Now substituting back into our integration by parts result:
\[ \int \tan^{-1}(x) \, dx = x \tan^{-1}(x) - \frac{1}{2} \ln(1+x^2) + C \]
Therefore,
\[ I = 3 \left( x \tan^{-1}(x) - \frac{1}{2} \ln(1+x^2) \right) + C \]
Simplifying gives:
\[ I = 3x \tan^{-1}(x) - \frac{3}{2} \ln(1+x^2) + C \]

### Final Answer:
Thus, the integral \( \int \tan^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right) dx \) is:
\[ I = 3x \tan^{-1}(x) - \frac{3}{2} \ln(1+x^2) + C \]