Question

Evaluate the integral: \[ \int \frac{x^2 + 1}{x^4 + 1} \, dx \] The correct answer is:

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

Hint:

When integrating expressions like \(\frac{x^2 + 1}{x^4 + 1}\), look for factoring patterns or use trigonometric identities to simplify the integrand.

Answer 1

The Correct Option is B

We are asked to evaluate the integral: \[ I = \int \frac{x^2 + 1}{x^4 + 1} \, dx. \] To solve this, notice that: \[ x^4 + 1 = (x^2 + 1)(x^2 - 1). \] Now, rewrite the integrand: \[ \frac{x^2 + 1}{x^4 + 1} = \frac{1}{x^2 + 1} - \frac{x}{x^2 + 1}. \] We can then separate the integral into two parts: \[ I = \int \frac{1}{x^2 + 1} \, dx - \int \frac{x}{x^2 + 1} \, dx. \] The first part is a standard integral: \[ \int \frac{1}{x^2 + 1} \, dx = \tan^{-1}(x). \] For the second part, use substitution: \( u = x^2 + 1 \), so \( du = 2x \, dx \). This gives: \[ \int \frac{x}{x^2 + 1} \, dx = \frac{1}{2} \ln|x^2 + 1| = \frac{1}{2} \ln(x^2 + 1). \] Thus, combining the two parts, we get: \[ I = \tan^{-1}(x) + \frac{1}{2} \ln(x^2 + 1) + c. \] Therefore, the correct answer is \( (B) \).