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:

Whenever an integral contains a function multiplied by its derivative: \[ \int f(x)f'(x)\,dx \] use substitution \(t=f(x)\). Then \[ \int f(x)f'(x)\,dx = \frac{f(x)^2}{2} + C \] This pattern appears frequently in inverse trigonometric integrals.

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).