Question

Evaluate the integral: \[ \int \frac{x^3 \tan^{-1}(x^4)}{1 + x^8} \,dx. \]

• A. \( \frac{(\tan^{-1}(x^4))^2}{8} + c \)
• B. \( \frac{(\tan^{-1}(x^4))^3}{3} + c \)
• C. \( \frac{(\tan^{-1}(x^4))^2}{4} + c \)
• D. \( \frac{(\tan^{-1}(x^4))^2}{2} + c \)

Hint:

For integration involving inverse trigonometric functions, check if differentiation of the inverse function appears in the numerator.

Answer 1

The Correct Option is A

Step 1: Substituting \( u = \tan^{-1}(x^4) \) Differentiate both sides: \[ du = \frac{4x^3}{1 + x^8} dx. \] Rearranging: \[ x^3 dx = \frac{(1 + x^8) du}{4}. \] Substituting in the given integral: \[ \int u \cdot \frac{(1 + x^8) du}{4}. \] \[ \frac{1}{4} \int u \, du. \]
Step 2: Solving the Integral \[ \frac{1}{4} \cdot \frac{u^2}{2} = \frac{u^2}{8}. \] \[ = \frac{(\tan^{-1}(x^4))^2}{8} + c. \] Thus, the correct answer is option (1).