BIBO stability of $y(t)=\int_{-\infty}^{t}{x(\tau)d\tau}$
Since this is most likely homework, here is a hint.

Write the integral you have displayed in the form $\int_{-\infty}^\infty x(\tau)h(t-\tau) d\tau$ where you get to choose what the function $h(\cdot)$ is to make it all work out. Then, $h(t)$ is the impulse response of the LTI system, Do you know the criterion for BIBO stability of an LTI system in terms of $h(t)$? Does your $h(t)$ satisfy the criterion?

Edit in response to OP's comment and definition of BIBO stability:

Your understanding of the definition of BIBO stability is faulty.


  A system with input $x(t)$ and output $y(t)$ is said to be bounded-input bounded-output (BIBO) stable if it has the property that whenever $x(t)$ is bounded (that is, there exists a (finite) real number $M$ such that $|x(t)| \leq M$ for all real numbers $t$), then the output is also 
  bounded (that is, there exists a (finite) real number $N$ (possibly dependent on $M$) such that $|y(t)| \leq N$ for all real numbers $t$).


Note that while $N$ might depend on $M$, it does not depend on $t$, the same $N$ must be an upper bound on $|y(t)|$ for all $t$.

In your case, with $y(t) = \int_{-\infty}^t x(\tau) d\tau$, what happens if $x(t) = u(t)$ for all real numbers $t$ (step input)? This is a bounded input. Does there exist a finite number $N$ such that $|y(t)| \leq N$ for all real numbers $t$?