Techniques of Integration w/ absolute value I cannot solve this integral.
$$
\int_{-3}^3 \frac{x}{1+|x|} ~dx
$$
I have tried rewriting it as:
$$
\int_{-3}^3 1 - \frac{1}{x+1}~dx,
$$
From which I obtain:
$$
x - \ln|x+1|\;\;\bigg\vert_{-3}^{\;3}
$$
My book (Stewart's Calculus 7e) has the answer as 0, and I can intuitively see this from a graph of the function, it being symmetric about $y = x$, but I cannot analytically solve this.

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. 
