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: Factorise the denominator
\(x^6+1 = (x^2+1)(x^4-x^2+1)\).

Step 2: Split the numerator
\(x^4+1 = (x^4-x^2+1) + x^2\). So \(\frac{x^4+1}{x^6+1} = \frac{1}{x^2+1} + \frac{x^2}{x^6+1}\).

Step 3: Integrate each
\(\int\frac{dx}{1+x^2} = \tan^{-1}x\). For the second, put \(u=x^3\), \(du = 3x^2dx\), giving \(\frac13\int\frac{du}{1+u^2} = \frac13\tan^{-1}(x^3)\).

Step 4: Result
\(\tan^{-1}x + \frac13\tan^{-1}(x^3)+c\), option (B).

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