Step 1: Understanding the Question:
We are asked to find the indefinite integral of a function that involves an inverse cotangent term. The structure of the integrand suggests that the method of substitution would be effective.
Step 2: Key Formula or Approach:
We should look for a function and its derivative. Let's consider the term \( u = \cot^{-1}(x^3) \). We need to find its derivative.
Recall the derivative formula for inverse cotangent: \( \frac{d}{du}(\cot^{-1}u) = -\frac{1}{1+u^2} \).
Using the chain rule, let \( u = x^3 \). Then \( \frac{du}{dx} = 3x^2 \).
So, \( \frac{d}{dx}(\cot^{-1}(x^3)) = -\frac{1}{1+(x^3)^2} \cdot \frac{d}{dx}(x^3) = -\frac{3x^2}{1+x^6} \).
The term \( \frac{x^2}{1+x^6} \) is present in the integrand, confirming that substitution is the correct method.
Step 3: Detailed Explanation:
1. Choose the substitution:
Let \( t = \cot^{-1}(x^3) \).
2. Find dt in terms of dx:
As calculated above, \( \frac{dt}{dx} = -\frac{3x^2}{1+x^6} \).
Rearranging this, we get: \( x^2 \,dx = -\frac{1+x^6}{3} \,dt \). Or more usefully, \( \frac{x^2}{1+x^6} \,dx = -\frac{dt}{3} \).
3. Rewrite the integral in terms of t:
Let's rearrange the original integral to group the terms for substitution:
\[ I = \int 4 \cdot \cot^{-1}(x^3) \cdot \frac{x^2}{1+x^6} \,dx \]
Now substitute \( t \) for \( \cot^{-1}(x^3) \) and \( -\frac{dt}{3} \) for \( \frac{x^2}{1+x^6} \,dx \).
\[ I = \int 4 \cdot t \cdot \left(-\frac{dt}{3}\right) = -\frac{4}{3} \int t \,dt \]
4. Integrate with respect to t:
Using the power rule for integration:
\[ I = -\frac{4}{3} \left( \frac{t^2}{2} \right) + C = -\frac{2}{3} t^2 + C \]
5. Substitute back to get the answer in terms of x:
Replace \( t \) with \( \cot^{-1}(x^3) \).
\[ I = -\frac{2}{3} (\cot^{-1}(x^3))^2 + C \]
Step 4: Final Answer:
The integral evaluates to \( -\dfrac{2}{3}(\cot^{-1}x^3)^2 + C \).