What is `int(x^(4) +1)/(x^(2) + 1)dx` equal to ?
Where 'c' is a constant of integration

Options

A. `(x^(3))/(3) - x + 4 tan^(-1) x + c`
B. `(x^(3))/(3) + x + 4 tan^(-1) x + c`
C. `(x^(3))/(3) - x + 2 tan^(-1) x + c`
D. `(x^(3))/(3) - x - 4 tan^(-1) x + c`

Explanation

To solve the integral \( \int \frac{x^4 + 1}{x^2 + 1} \, dx \), we can break it down step by step.

### Step 1: Simplify the integrand
We start by rewriting the integrand:
\[ \frac{x^4 + 1}{x^2 + 1} = \frac{x^4 + 2 - 1}{x^2 + 1} = \frac{x^4 - 1}{x^2 + 1} + \frac{2}{x^2 + 1} \]
This allows us to separate the integral into two parts:
\[ \int \frac{x^4 + 1}{x^2 + 1} \, dx = \int \frac{x^4 - 1}{x^2 + 1} \, dx + \int \frac{2}{x^2 + 1} \, dx \]

### Step 2: Factor the numerator
Next, we can factor \( x^4 - 1 \):
\[ x^4 - 1 = (x^2 - 1)(x^2 + 1) \]
Thus, we can rewrite the first integral:
\[ \int \frac{x^4 - 1}{x^2 + 1} \, dx = \int \frac{(x^2 - 1)(x^2 + 1)}{x^2 + 1} \, dx = \int (x^2 - 1) \, dx \]

### Step 3: Integrate the first part
Now we can integrate \( x^2 - 1 \):
\[ \int (x^2 - 1) \, dx = \frac{x^3}{3} - x \]

### Step 4: Integrate the second part
Now we integrate the second part:
\[ \int \frac{2}{x^2 + 1} \, dx = 2 \int \frac{1}{x^2 + 1} \, dx = 2 \tan^{-1}(x) \]

### Step 5: Combine the results
Putting it all together, we have:
\[ \int \frac{x^4 + 1}{x^2 + 1} \, dx = \left( \frac{x^3}{3} - x \right) + 2 \tan^{-1}(x) + C \]

### Final Answer
Thus, the final result is:
\[ \int \frac{x^4 + 1}{x^2 + 1} \, dx = \frac{x^3}{3} - x + 2 \tan^{-1}(x) + C \]