Time Variant or Time-Invariant system?
The system is not time invariant as can be seen by the fundamental test procudere:

Given the system with an input/output relation:
$$y(t)=\mathcal{T} \{ x(t) \} = \int_{-\infty}^{x(t)} x(\tau) d\tau$$

First check the new output for a shifted input $x(t-d)$ as
$$y_d(t) = \mathcal{T} \{ x(t-d) \} = \int_{-\infty}^{x(t-d)} x(\tau-d) d\tau $$
$$y_d(t) = \mathcal{T} \{ x(t-d) \} = \int_{-\infty}^{x(t-d)-d} x(\tau) d\tau $$

And then also check if the output $y(t)$ shifted by $d$ yields th same formula?
We know that output $y(t)$ is given in terms of input $x(t)$ with the formula
$$y(t) = \int_{-\infty}^{x(t)} x(\tau) d\tau$$

Hence the shifted output $y(t-d)$ is obtained by simply replacing every $t$ variable with $t-d$ in that formula which yields: 
$$y(t-d) = \int_{-\infty}^{x(t-d)} x(\tau) d\tau $$

And now looking at the formulas for both $y(t-d)$ and $\mathcal{T} \{ x(t-d) \}$ given above, we see that since $x(t-d)-d \neq x(t-d)$ for all $d$ then $y(t-d) \neq y_d(t)$ and we conclude that the system is not time invariant.