Question

The integral \( \int \frac{2+x^4}{1+x^2} dx \) is equal to

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

Hint:

When integrating rational functions where the numerator's degree is greater than or equal to the denominator's degree, always perform algebraic division (or manipulation) first. This breaks the complex fraction into simpler terms (a polynomial and a proper rational fraction) that are easier to integrate.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks to evaluate an indefinite integral of a rational function. The degree of the numerator is higher than the degree of the denominator, so algebraic manipulation or polynomial long division is required before integration.

Step 2: Key Formula or Approach:
1. Perform algebraic manipulation to simplify the integrand: \[ \frac{x^4+2}{x^2+1} = \frac{x^4-1+3}{x^2+1} \] 2. Use the algebraic identity \(a^2 - b^2 = (a-b)(a+b)\) for \(x^4-1 = (x^2-1)(x^2+1)\).
3. Apply standard integral formulas: \[ \int x^n dx = \frac{x^{n+1}}{n+1} + C \quad (\text{for } n \neq -1) \] \[ \int \frac{1}{1+x^2} dx = \tan^{-1}x + C \]

Step 3: Detailed Explanation:
Given integral: \( I = \int \frac{2+x^4}{1+x^2} dx \)
First, manipulate the numerator to simplify the fraction:
\[ I = \int \frac{x^4 - 1 + 3}{x^2 + 1} dx \]
Split the fraction into two terms:
\[ I = \int \left( \frac{x^4 - 1}{x^2 + 1} + \frac{3}{x^2 + 1} \right) dx \]
Factor the numerator of the first term using the difference of squares: \( x^4 - 1 = (x^2 - 1)(x^2 + 1) \).
\[ I = \int \left( \frac{(x^2 - 1)(x^2 + 1)}{x^2 + 1} + \frac{3}{x^2 + 1} \right) dx \]
Simplify the first term:
\[ I = \int \left( (x^2 - 1) + \frac{3}{x^2 + 1} \right) dx \]
Now, integrate each term separately:
\[ I = \int x^2 dx - \int 1 dx + 3 \int \frac{1}{x^2 + 1} dx \]
Apply the standard integral formulas:
\[ I = \frac{x^{2+1}}{2+1} - x + 3 \tan^{-1}x + C \]
\[ I = \frac{x^3}{3} - x + 3 \tan^{-1}x + C \]
This result matches option (a).

Step 4: Final Answer:
The integral is equal to \(\frac{1}{3}x^3 + 3 \tan^{-1}x - x + C\).