` int _(-1)^(1)(x^(3)+|x|+1)/(x^(2)+2|x|+1)dx` is equal to

Options

A. log 2
B. 2 log 2
C. `1/2 log2`
D. `4 log 2`

Explanation

To evaluate the integral

\[ I = \int_{-1}^{1} \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1} \, dx, \]

we can break this integral into two parts based on the properties of the absolute value function.

### Step 1: Split the Integral
We can split the integral into two parts, one from \(-1\) to \(0\) and the other from \(0\) to \(1\):

\[ I = \int_{-1}^{0} \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1} \, dx + \int_{0}^{1} \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1} \, dx. \]

### Step 2: Evaluate the Integral from \(-1\) to \(0\)
For \(x\) in the interval \([-1, 0]\), we have \(|x| = -x\). Thus, the integral becomes:

\[ \int_{-1}^{0} \frac{x^3 - x + 1}{x^2 - 2x + 1} \, dx. \]

The denominator simplifies to:

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

So we can rewrite the integral as:

\[ \int_{-1}^{0} \frac{x^3 - x + 1}{(x-1)^2} \, dx. \]

### Step 3: Evaluate the Integral from \(0\) to \(1\)
For \(x\) in the interval \([0, 1]\), we have \(|x| = x\). Thus, the integral becomes:

\[ \int_{0}^{1} \frac{x^3 + x + 1}{x^2 + 2x + 1} \, dx. \]

The denominator simplifies to:

\[ x^2 + 2x + 1 = (x+1)^2. \]

So we can rewrite the integral as:

\[ \int_{0}^{1} \frac{x^3 + x + 1}{(x+1)^2} \, dx. \]

### Step 4: Combine the Results
Now we have:

\[ I = \int_{-1}^{0} \frac{x^3 - x + 1}{(x-1)^2} \, dx + \int_{0}^{1} \frac{x^3 + x + 1}{(x+1)^2} \, dx. \]

### Step 5: Analyze the Functions
Notice that the first integral is an odd function and the second integral is an even function. The integral of an odd function over a symmetric interval around zero is zero:

\[ \int_{-1}^{0} \frac{x^3 - x + 1}{(x-1)^2} \, dx = 0. \]

Thus, we only need to evaluate the second integral:

\[ I = \int_{0}^{1} \frac{x^3 + x + 1}{(x+1)^2} \, dx. \]

### Step 6: Evaluate the Integral
Now we can compute:

\[ \int_{0}^{1} \frac{x^3 + x + 1}{(x+1)^2} \, dx. \]

This can be done by polynomial long division or by using substitution.

### Final Calculation
After evaluating the integral, we find that:

\[ I = 2. \]

### Conclusion
Thus, the value of the integral is:

\[ \int_{-1}^{1} \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1} \, dx = 2. \]