We need to evaluate the definite integral $I = \int_{-1}^{1} \frac{x}{1 + |x|} \, dx$.

1. Problem Description

We need to evaluate the definite integral $I = \int_{-1}^{1} \frac{x}{1 + |x|} \, dx$.

2. Solution Steps

Since the integral is from $-1$ to $1$, we can split the integral into two parts based on the sign of $x$.

$I = \int_{-1}^{0} \frac{x}{1 + |x|} \, dx + \int_{0}^{1} \frac{x}{1 + |x|} \, dx$

When $x < 0$, $|x| = -x$.

When $x > 0$, $|x| = x$.

So,

$I = \int_{-1}^{0} \frac{x}{1 - x} \, dx + \int_{0}^{1} \frac{x}{1 + x} \, dx$

Let's consider the first integral, $I_1 = \int_{-1}^{0} \frac{x}{1 - x} \, dx$. We can rewrite the integrand as:

$\frac{x}{1 - x} = \frac{-(1-x)+1}{1-x} = -1 + \frac{1}{1 - x}$.

Then $I_1 = \int_{-1}^{0} \left(-1 + \frac{1}{1 - x}\right) \, dx = \left[-x - \ln|1 - x|\right]_{-1}^{0} = (0 - \ln|1|) - (1 - \ln|1 - (-1)|) = 0 - (1 - \ln 2) = -1 + \ln 2$.

Now consider the second integral, $I_2 = \int_{0}^{1} \frac{x}{1 + x} \, dx$. We can rewrite the integrand as:

$\frac{x}{1 + x} = \frac{(1+x)-1}{1+x} = 1 - \frac{1}{1 + x}$.

Then $I_2 = \int_{0}^{1} \left(1 - \frac{1}{1 + x}\right) \, dx = \left[x - \ln|1 + x|\right]_{0}^{1} = (1 - \ln|1 + 1|) - (0 - \ln|1 + 0|) = 1 - \ln 2 - 0 = 1 - \ln 2$.

So, $I = I_1 + I_2 = (-1 + \ln 2) + (1 - \ln 2) = 0$.

3. Final Answer

0