Evaluate: `int(x^2tan^(-1)x^3)/(1+x^6)dx`

Explanation

To evaluate the integral

\[ \int \frac{x^2 \tan^{-1}(x^3)}{1+x^6} \, dx, \]

we can use a substitution method. Let's go through the steps:

### Step 1: Substitution
Let

\[ t = \tan^{-1}(x^3). \]

Then, we need to find \(dt\). The derivative of \(t\) with respect to \(x\) is:

\[ \frac{dt}{dx} = \frac{1}{1 + (x^3)^2} \cdot \frac{d}{dx}(x^3) = \frac{3x^2}{1 + x^6}. \]

Thus, we have:

\[ dt = \frac{3x^2}{1 + x^6} \, dx. \]

Rearranging gives:

\[ dx = \frac{1 + x^6}{3x^2} \, dt. \]

### Step 2: Substitute in the Integral
Now we substitute \(t\) and \(dx\) into the integral:

\[ \int \frac{x^2 \tan^{-1}(x^3)}{1+x^6} \, dx = \int \frac{x^2 t}{1+x^6} \cdot \frac{1+x^6}{3x^2} \, dt. \]

The \(x^2\) and \(1+x^6\) terms cancel out:

\[ = \frac{1}{3} \int t \, dt. \]

### Step 3: Integrate
Now we can integrate:

\[ \frac{1}{3} \int t \, dt = \frac{1}{3} \cdot \frac{t^2}{2} + C = \frac{t^2}{6} + C. \]

### Step 4: Substitute Back
Now we substitute back \(t = \tan^{-1}(x^3)\):

\[ = \frac{(\tan^{-1}(x^3))^2}{6} + C. \]

### Final Answer
Thus, the final result of the integral is:

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

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