What is `int(x^(4)+1)/(x^(2)+1)dx` equal to ?

Options

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

Answer

To solve the integral \( \int \frac{x^4 + 1}{x^2 + 1} \, dx \), we can follow these steps:

### Step 1: Simplify the integrand
We start with the expression:

\[ \frac{x^4 + 1}{x^2 + 1} \]

We can rewrite \( x^4 + 1 \) as \( (x^4 - 1) + 2 \). Thus, we have:

\[ x^4 + 1 = (x^4 - 1) + 2 = (x^2 - 1)(x^2 + 1) + 2 \]

So, we can express the integral as:

\[ \int \frac{(x^2 - 1)(x^2 + 1) + 2}{x^2 + 1} \, dx \]

### Step 2: Split the integral
Now we can split the integral into two parts:

\[ \int \frac{(x^2 - 1)(x^2 + 1)}{x^2 + 1} \, dx + \int \frac{2}{x^2 + 1} \, dx \]

This simplifies to:

\[ \int (x^2 - 1) \, dx + 2 \int \frac{1}{x^2 + 1} \, dx \]

### Step 3: Solve the first integral
Now we solve the first integral:

\[ \int (x^2 - 1) \, dx = \int x^2 \, dx - \int 1 \, dx \]

Using the power rule for integration:

\[ \int x^2 \, dx = \frac{x^3}{3} \quad \text{and} \quad \int 1 \, dx = x \]

So, we have:

\[ \int (x^2 - 1) \, dx = \frac{x^3}{3} - x \]

### Step 4: Solve the second integral
Next, we solve the second integral:

\[ 2 \int \frac{1}{x^2 + 1} \, dx \]

The integral of \( \frac{1}{x^2 + 1} \) is known to be:

\[ \int \frac{1}{x^2 + 1} \, dx = \tan^{-1}(x) \]

Thus, we have:

\[ 2 \int \frac{1}{x^2 + 1} \, dx = 2 \tan^{-1}(x) \]

### Step 5: Combine the results
Now, we combine the results of both integrals:

\[ \int \frac{x^4 + 1}{x^2 + 1} \, dx = \left( \frac{x^3}{3} - x \right) + 2 \tan^{-1}(x) + C \]

### Final Answer
So, the final result is:

\[ \int \frac{x^4 + 1}{x^2 + 1} \, dx = \frac{x^3}{3} - x + 2 \tan^{-1}(x) + C \]