Step 1: Understanding the Concept:
When integrating rational functions where the degree of the numerator is greater than or equal to the degree of the denominator, the function is called an "improper fraction."
The standard procedure is to transform it into a proper form using polynomial division or algebraic manipulation.
In this specific case, the numerator is of degree 4 and the denominator is of degree 2.
Step 2: Key Formula or Approach:
We will use the technique of "adding and subtracting" terms in the numerator to create terms divisible by the denominator.
Useful standard integrals:
1. \( \int x^{n} dx = \frac{x^{n+1}}{n+1} + C \)
2. \( \int \frac{1}{1+x^{2}} dx = \tan^{-1} x + C \)
3. \( \int 1 dx = x + C \)
Step 3: Detailed Explanation:
Consider the integral \( I = \int \frac{x^{4} + 2}{x^{2} + 1} dx \).
We know that \( x^{4} - 1 \) is divisible by \( x^{2} + 1 \) because it follows the identity \( a^{2} - b^{2} = (a - b)(a + b) \).
Specifically, \( x^{4} - 1 = (x^{2})^{2} - 1^{2} = (x^{2} - 1)(x^{2} + 1) \).
Let's rewrite the numerator:
\[ x^{4} + 2 = (x^{4} - 1) + 3 \]
Now, substitute this into the integral:
\[ I = \int \frac{(x^{4} - 1) + 3}{x^{2} + 1} dx \]
Split the integral into two parts:
\[ I = \int \frac{x^{4} - 1}{x^{2} + 1} dx + \int \frac{3}{x^{2} + 1} dx \]
Simplify the first fraction by canceling the common factor \( (x^{2} + 1) \):
\[ \frac{(x^{2} - 1)(x^{2} + 1)}{x^{2} + 1} = x^{2} - 1 \]
The expression now becomes:
\[ I = \int (x^{2} - 1) dx + 3 \int \frac{1}{1+x^{2}} dx \]
Apply the integration rules term by term:
\[ I = \left( \frac{x^{3}}{3} - x \right) + 3 \tan^{-1} x + C \]
Rearrange to match the format of the given options:
\[ I = \frac{1}{3}x^{3} + 3 \tan^{-1} x - x + C \]
Verification: Differentiating the answer gives \( x^{2} - 1 + \frac{3}{1+x^{2}} = \frac{(x^{2}-1)(1+x^{2}) + 3}{1+x^{2}} = \frac{x^{4}-1+3}{1+x^{2}} = \frac{x^{4}+2}{1+x^{2}} \), which is the original integrand.
Step 4: Final Answer:
The integral evaluates to \( \frac{1}{3}x^{3} + 3 \tan^{-1} x - x + C \).