`int(3x^(2)+2x)/(x^(6)+2x^(5)+x^(4)+2x^(3)+2x^(2)+5)dx=`

Options

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

Explanation

To solve the integral

\[ \int \frac{3x^2 + 2x}{x^6 + 2x^5 + x^4 + 2x^3 + 2x^2 + 5} \, dx, \]

we can follow these steps:

### Step 1: Identify the Denominator and its Derivative
We start by observing the denominator:

\[ x^6 + 2x^5 + x^4 + 2x^3 + 2x^2 + 5. \]

Notice that the derivative of the denominator resembles the numerator. We can differentiate the denominator to see if it matches the numerator:

\[ \frac{d}{dx}(x^6 + 2x^5 + x^4 + 2x^3 + 2x^2 + 5) = 6x^5 + 10x^4 + 4x^3 + 6x^2 + 4x. \]

### Step 2: Simplify the Integral
Now, we can rewrite the integral in a more manageable form. We can express the numerator in terms of the derivative of the denominator:

\[ 3x^2 + 2x = \frac{d}{dx}(x^3 + x^2 + 1). \]

### Step 3: Use Substitution
Let

\[ t = x^3 + x^2 + 1. \]

Then, the derivative \(dt\) is:

\[ dt = (3x^2 + 2x) \, dx. \]

This allows us to rewrite the integral as:

\[ \int \frac{dt}{t^2 + 4}. \]

### Step 4: Recognize the Standard Integral Form
The integral

\[ \int \frac{dt}{t^2 + 4} \]

is a standard integral that can be solved using the formula:

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

In our case, \(a^2 = 4\) so \(a = 2\):

\[ \int \frac{dt}{t^2 + 4} = \frac{1}{2} \tan^{-1} \left( \frac{t}{2} \right) + C. \]

### Step 5: Substitute Back
Now, we substitute back for \(t\):

\[ \frac{1}{2} \tan^{-1} \left( \frac{x^3 + x^2 + 1}{2} \right) + C. \]

### Final Answer
Thus, the final solution to the integral is:

\[ \int \frac{3x^2 + 2x}{x^6 + 2x^5 + x^4 + 2x^3 + 2x^2 + 5} \, dx = \frac{1}{2} \tan^{-1} \left( \frac{x^3 + x^2 + 1}{2} \right) + C. \]