Question

Evaluate the integral \[ \int \frac{x^4 + 1}{x^6 + 1} dx. \]

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

Hint:

For rational functions, factor and use partial fraction decomposition.

Answer 1

The Correct Option is D

Step 1: Splitting the fraction \[ I = \int \frac{x^4 + 1}{x^6 + 1} dx. \] Factoring: \[ x^6 + 1 = (x^2 + 1)(x^4 - x^2 + 1). \] Step 2: Using partial fractions \[ \frac{x^4 + 1}{x^6 + 1} = \frac{A}{x^2 + 1} + \frac{B}{x^4 - x^2 + 1}. \] Solving for coefficients and integrating leads to: \[ I = \tan^{-1} x + \frac{1}{3} \tan^{-1} x^3 + c. \]

Answer 2

Evaluate the integral:
\[ I = \int \frac{x^4 + 1}{x^6 + 1} \, dx \]

Step 1: Factor the denominator:
\[ x^6 + 1 = (x^2 + 1)(x^4 - x^2 + 1) \]

Step 2: Use partial fraction decomposition:
Assume:
\[ \frac{x^4 + 1}{(x^2 + 1)(x^4 - x^2 + 1)} = \frac{A x^2 + B}{x^2 + 1} + \frac{C x^3 + D x^2 + E x + F}{x^4 - x^2 + 1} \] However, since the degree of numerator is less than denominator, and numerator is symmetric, try the substitution method.

Step 3: Use substitution \( t = x^3 \), then \( dt = 3x^2 dx \), or try rewriting numerator:
Rewrite numerator:
\[ x^4 + 1 = x^4 - x^2 + 1 + x^2 \] So, \[ \frac{x^4 + 1}{x^6 + 1} = \frac{x^4 - x^2 + 1 + x^2}{(x^2 + 1)(x^4 - x^2 + 1)} = \frac{1}{x^2 + 1} + \frac{x^2}{(x^2 + 1)(x^4 - x^2 + 1)} \] but this is complicated.

Step 4: Use the identity:
\[ \frac{d}{dx} \tan^{-1} x = \frac{1}{1 + x^2} \] \[ \frac{d}{dx} \tan^{-1} x^3 = \frac{3x^2}{1 + x^6} \] Rewrite the integral as:
\[ I = \int \frac{1}{1 + x^2} dx + \frac{1}{3} \int \frac{3x^2}{1 + x^6} dx \] But this matches the derivatives of \( \tan^{-1} x \) and \( \tan^{-1} x^3 \).

Step 5: Therefore,
\[ I = \tan^{-1} x + \frac{1}{3} \tan^{-1} x^3 + C \]

Hence, the integral evaluates to:
\[ \boxed{\tan^{-1} x + \frac{1}{3} \tan^{-1} x^3 + C} \]