Evaluate `int_(-1)^(1) (x+|x|+1)/(x^(2)+2|x|+1)`

Answer

To evaluate the integral

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

we will break it down step by step.

### Step 1: Analyze the integrand

The expression \( |x| \) behaves differently based on the sign of \( x \):
- For \( x < 0 \): \( |x| = -x \)
- For \( x \geq 0 \): \( |x| = x \)

Thus, we can split the integral into two parts:

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

### Step 2: Simplify the first integral

For \( x \in [-1, 0] \):

\[ I_1 = \int_{-1}^{0} \frac{1}{(x - 1)^2} \, dx. \]

The denominator simplifies to \( (x + 1)^2 \) because \( x^2 - 2(-x) + 1 = x^2 + 2x + 1 \).

### Step 3: Simplify the second integral

For \( x \in [0, 1] \):

\[ I_2 = \int_{0}^{1} \frac{2x + 1}{(x + 1)^2} \, dx. \]

### Step 4: Evaluate the first integral

Now, we evaluate \( I_1 \):

\[ I_1 = \int_{-1}^{0} \frac{1}{(x + 1)^2} \, dx. \]

This integral can be solved using the formula for the integral of \( \frac{1}{x^n} \):

\[ I_1 = \left[-\frac{1}{x + 1}\right]_{-1}^{0} = -\frac{1}{0 + 1} + \frac{1}{-1 + 1} = -1 + \infty. \]

### Step 5: Evaluate the second integral

Now, we evaluate \( I_2 \):

\[ I_2 = \int_{0}^{1} \frac{2x + 1}{(x + 1)^2} \, dx. \]

This can be split into two parts:

\[ I_2 = \int_{0}^{1} \frac{2x}{(x + 1)^2} \, dx + \int_{0}^{1} \frac{1}{(x + 1)^2} \, dx. \]

### Step 6: Solve \( I_2 \)

1. For \( \int_{0}^{1} \frac{2x}{(x + 1)^2} \, dx \), use substitution or integration by parts.
2. For \( \int_{0}^{1} \frac{1}{(x + 1)^2} \, dx \):

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

Evaluating from 0 to 1 gives:

\[ \left[-\frac{1}{x + 1}\right]_{0}^{1} = -\frac{1}{2} + 1 = \frac{1}{2}. \]

### Step 7: Combine results

Since \( I_1 \) diverges and \( I_2 \) converges, the overall integral \( I \) diverges.

Thus, the final result is:

\[ I = \text{Diverges}. \]