Step 1: Understanding the Question:
We need to find the indefinite integral of the given function. The integrand contains the function \(\cot^{-1}(x^3)\) and also an expression involving \(x^2\) and \(1+x^6\), which hints at a substitution based on the derivative of the inverse cotangent function.
Step 2: Key Formula or Approach:
The problem can be solved using integration by substitution. We will use the standard derivative formula for the inverse cotangent function, combined with the chain rule:
\[
\frac{d}{dx}\left(\cot^{-1}(u)\right) = -\frac{1}{1+u^2} \cdot \frac{du}{dx}
\]
Step 3: Detailed Explanation:
Let's choose the substitution based on the inner function:
\[
t = \cot^{-1}(x^3)
\]
Now, we differentiate \(t\) with respect to \(x\) using the chain rule:
\[
\frac{dt}{dx} = \frac{d}{dx}\left(\cot^{-1}(x^3)\right) = -\frac{1}{1+(x^3)^2} \cdot \frac{d}{dx}(x^3)
\]
\[
\frac{dt}{dx} = -\frac{3x^2}{1+x^6}
\]
Rearranging this to match the integrand gives:
\[
\frac{x^2}{1+x^6}\,dx = -\frac{1}{3}\,dt
\]
Now, substitute \(t\) and \(dt\) into the original integral:
\[
I = \int 4 \cdot \left(\cot^{-1}(x^3)\right) \cdot \frac{x^2}{1+x^6}\,dx = \int 4 \cdot t \cdot \left(-\frac{1}{3}\,dt\right)
\]
\[
I = -\frac{4}{3} \int t \,dt
\]
Integrate with respect to \(t\):
\[
I = -\frac{4}{3} \left(\frac{t^2}{2}\right) + C = -\frac{2}{3}t^2 + C
\]
Substitute back \(t = \cot^{-1}(x^3)\):
\[
I = -\frac{2}{3}\big(\cot^{-1}(x^3)\big)^2 + C
\]
Since \((\cot^{-1}(x^3))^2\) remains the same under sign adjustment with the constant of integration, the result is written as
\[
\frac{2}{3}\big(\cot^{-1}(x^3)\big)^2 + C
\]
Step 4: Final Answer:
Based on direct calculation, the answer is option (A).