You have an odd function integrated over an interval symmetric about the origin. The answer is $0$. This is, as you point out, intuitively reasonable from the geometry. A general analytic argument is given in a remark at the end.
If you really want to calculate, break up the region into two parts, where (i) $x$ is positive and (ii) where $x$ is negative. For $x$ positive, you are integrating $\frac{x}{1+x}$. For $x$ negative, you are integrating $\frac{x}{1-x}$, since for $x\le 0$ we have $|x|=-x$. Finally, calculate $\int_{3}^0 \frac{x}{1-x}\,dx\quad\text{and}\quad\int_0^3 \frac{x}{1+x}\,dx$ and add. All that work for nothing!
Remarks: $1.$ In general, when absolute values are being integrated, breaking up into appropriate parts is the safe way to go. Trying to handle both at the same time carries too high a probability of error.
$2.$ Let $f(x)$ be an odd function. We want to integrate from $-a$ to $a$, where $a$ is positive. We look at $\int_{-a}^0 f(x)\,dx.$ Let $u=-x$. Then $du=-dx$, and $f(u)=-f(x)$. So our integral is $\int_{u=a}^0 (-1)(-f(u))\,du.$ the two minus signs cancel. Changing the order of the limits, we get $-\int_0^a f(u)\,du$. so this is just the negative of the integral over the interval from $0$ to $a$. That gives us the desired cancellation.