Question

\(\int \frac{x^4+1}{x^6+1}dx =\)

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

Hint:

Write x^6+1 as (x^2+1)(x^4-x^2+1) and split the numerator.

Answer 1

The Correct Option is B

Step 1: Differentiate option B
$\frac{d}{dx}\left[\tan^{-1}x+\frac13\tan^{-1}x^3\right] = \frac1{1+x^2} + \frac{x^2}{1+x^6}$.

Step 2: Combine
$\frac{(1+x^6) + x^2(1+x^2)}{(1+x^2)(1+x^6)} = \frac{x^6+x^4+x^2+1}{(1+x^2)(1+x^6)}$. Note $x^6+x^4+x^2+1 = (x^2+1)(x^4+1)$.

Step 3: Simplify
So the derivative is $\frac{x^4+1}{1+x^6}$, the original integrand. Option (B).

Final Answer:
Option B. \[ \boxed{\text{(B)}\ \tan^{-1}x+\frac13\tan^{-1}(x^3)+c} \]