Splitting a tensor Is it possible to write $$\int d^3x \,\,\, x_i\,\,x_j\,\,\,f(\vec x)$$ where $f(\vec x)$ is some function of the position and the indices indicate which component, 
as a sum of a traceless tensor and a multiple of $\delta_{ij}$? If so, is there a good way of seeing how that may be achieved? 
Thank you.

I don't think the following point is relevant, but just in case, I should add that $f(\vec x)$ decays as $|\vec x|\to \infty$
 A: Define $$ M_{ij} = \int d^3x \, x_ix_jf(\vec x).$$
To make $M$ traceless, we subtract the appropriate multiple of identity,
$$N_{ij} = M_{ij}  - \frac{\mathop{\rm tr M}}3 \delta_{ij}.$$
It is easy to check that $\mathop{\rm tr N}=0$ and 
$$M_{ij} = N_{ij} + \frac{\mathop{\rm tr M}}3 \delta_{ij}$$
by definition.
A: I take it $f$ is a scalar field.  Then what you have is a linear operator $\underline T$ such that
$$x_i x_j f = \underline T(e_i;x) \cdot e_j \implies \underline T(a;x) = (a \cdot x) x f(x)$$
where $\underline T$ is linear in $a$ (but not $x$).
This operator is symmetric.  See that the adjoint is
$$\overline T(b) = (b \cdot x) x f(x) = \underline T(b)$$
Can you make this into the sum of a traceless operator and the identity?  Well, sure.  Let's find the trace, $T$.
$$T = \nabla_a \cdot \underline T(a) = \nabla_a \cdot (a \cdot x) x f(x) = x^2 f(x)$$
Now you can construct a tracefree operator $\underline F$ by subtracting out $Ta/3$.
$$\underline F(a) = \underline T(a) - Ta/3 = (a \cdot x) x f(x) - ax^2 f(x)/3$$
The part with nonzero trace is the part we subtracted out, $Ta/3 = T \underline I(a)/3$.  Remember, the Kronecker delta, $\delta_{ij}$, merely represents the components of the identity operator.
I leave it to you to evaluate the components $F_{ij}$ of the operator $\underline F$ now.  In addition, while it was possible to make the integrand into a tensor field, making the whole integral into a non-field tensor is harder to do without some information about the integral itself.
At any rate, the key you should take away here is that you can always construct a tracefree linear operator just by subtracting out the identity operator multiplied by the trace.
