Make ##d\xi/d\lambda = (d\xi^{\mu}/dx^{\mu})*dx^{\mu}/d\lambda##. Which will allow you to put the 2 ##\dot{x}## in evidence. ( this is the chain rule with ##x^{\mu}## until I find out to write out formulas in latex here)
Then you'll get ##2* \nabla \cdot \xi - \Omega^2## in the integrand.
There...