Question

Evaluate the integral: \[ \int \frac{4x^2 \cot^{-1}(x^3)}{1+x^6}\,dx \] (where \(C\) is a constant of integration)

• A. \( -\dfrac{2}{3}(\cot^{-1}x^3) + C \)
• B. \( \dfrac{2}{3}(\cot^{-1}x^3) + C \)
• C. \( -\dfrac{2}{3}(\cot^{-1}x^3)^2 + C \)
• D. \( \dfrac{2}{3}(\cot^{-1}x^3)^2 + C \)

Hint:

Before setting up a substitution, try differentiating cot inverse of x cubed using the chain rule and compare the result to the x squared term in the numerator of the integrand. If the given expression turns out to be a constant multiple of that derivative, you can write the antiderivative directly. Keep track of the constant multiplier carefully so it is not lost in the final answer.

Answer 1

The Correct Option is C

Concept: Use the substitution method when the integrand contains a function and its derivative. Recall the derivative: \[ \frac{d}{dx}(\cot^{-1}x) = -\frac{1}{1+x^2} \] For a composite function: \[ \frac{d}{dx}(\cot^{-1}(x^3)) = -\frac{3x^2}{1+x^6} \]
Step 1: {Choose substitution.} Let \[ t = \cot^{-1}(x^3) \] Then \[ \frac{dt}{dx} = -\frac{3x^2}{1+x^6} \] \[ dt = -\frac{3x^2}{1+x^6}\,dx \]
Step 2: {Rewrite the given integral.} \[ \int \frac{4x^2 \cot^{-1}(x^3)}{1+x^6}dx \] Using substitution: \[ \frac{4x^2}{1+x^6}dx = -\frac{4}{3}dt \] Thus the integral becomes \[ \int t\left(-\frac{4}{3}\right)dt \]
Step 3: {Integrate.} \[ -\frac{4}{3}\int t\,dt \] \[ = -\frac{4}{3}\cdot \frac{t^2}{2} \] \[ = -\frac{2}{3}t^2 \]
Step 4: {Substitute back \(t\).} \[ t = \cot^{-1}(x^3) \] Therefore, \[ \int \frac{4x^2 \cot^{-1}(x^3)}{1+x^6}\,dx = -\frac{2}{3}(\cot^{-1}(x^3))^2 + C \] Hence, the correct option is (C).

Answer 2

Concept:

• If an integrand can be written as a constant times the derivative of some function raised to a power, the antiderivative can be written down directly by reversing the chain rule, without introducing a new substitution variable.
• Key derivative needed here: $\dfrac{d}{dx}\left[\cot^{-1}(x^3)\right] = -\dfrac{3x^2}{1+x^6}$.

Step 1: Differentiate $\left[\cot^{-1}(x^3)\right]^2$ using the chain rule.
$\dfrac{d}{dx}\left[\left(\cot^{-1}(x^3)\right)^2\right] = 2\cot^{-1}(x^3) \cdot \dfrac{d}{dx}\left[\cot^{-1}(x^3)\right] = 2\cot^{-1}(x^3) \cdot \left(-\dfrac{3x^2}{1+x^6}\right) = -\dfrac{6x^2\cot^{-1}(x^3)}{1+x^6}$

Step 2: Compare this with the given integrand.
The integrand is $\dfrac{4x^2\cot^{-1}(x^3)}{1+x^6}$, which is $-\dfrac{2}{3}$ times the derivative found in Step 1:
$\dfrac{4x^2\cot^{-1}(x^3)}{1+x^6} = -\dfrac{2}{3}\left[-\dfrac{6x^2\cot^{-1}(x^3)}{1+x^6}\right] = -\dfrac{2}{3}\cdot \dfrac{d}{dx}\left[\left(\cot^{-1}(x^3)\right)^2\right]$

Step 3: Integrate both sides directly.
$\int \dfrac{4x^2\cot^{-1}(x^3)}{1+x^6}\,dx = -\dfrac{2}{3}\int \dfrac{d}{dx}\left[\left(\cot^{-1}(x^3)\right)^2\right]dx = -\dfrac{2}{3}\left(\cot^{-1}(x^3)\right)^2 + C$

Final Answer: $-\dfrac{2}{3}\left(\cot^{-1}x^3\right)^2 + C$