Is this system time invariant?
Hint

You are on the right track. You note that the integral is 


  not a normal convolution integral


and 


  there is no unit step $u(t)$ clearly defined in the transfer function. 


So, can you introduce a unit step yourself and change the limits of integration to make it look like a normal convolution integral?

STOP! Full answer below.

Let $\mathbf{1}_{A}$ be the indicator function which is 1 on the set $A$ and 0
everywhere else.
\begin{eqnarray}
y(t) &amp;=&amp; \int_{-\infty}^t e^{-9(t-\tau)} x(\tau) d\tau \\
&amp;=&amp; \int_{-\infty}^{\infty} e^{-9(t-\tau)} x(\tau) \mathbf{1}_{\{\tau \leq t\}} d\tau \\
&amp;=&amp; \int_{-\infty}^{\infty} e^{-9(t-\tau)} u(t-\tau) x(\tau) d\tau
\end{eqnarray}

This is an LTI system with transfer function $e^{-9t}u(t)$.