Okay, I felt like clarifying this.
The first one we could say, correctly, is 3t^2+C, where C is an arbitrary constant. We could also say, correctly, that it's \left(3t^2+1\right)+C, where C is an arbitrary constant. In fact, the point of the arbitrary constant is that 3t^2, 3t^2+1, 3t^2+2, and 3t^2\cdot\lim\limits_{a\to0^+}^{}\left( \displaystyle\int_a^1 \left(x^x\right)\cdot\mathrm{d}x\right) are all equally valid antiderivatives (or indefinite integrals.) 3t^2 isn't "special" in being the "fundamental" one which all others differ by a constant from. When we say "indefinite integral," we don't mean "integral with lower bound 0." We mean "integral from some arbitrary lower bound." Our C is going to depend on this lower bound (if we choose it to be 0, our C is 0, but this relation is false for integrating most functions.) It's probably best to say that \displaystyle\int\left(6t\right)\cdot\mathrm{d}t is the set of all functions f such that, for all x, f\left(x\right)=3x^2+C, where C is an arbitrary constant. (Again, this is also the set of all functions f such that, for all x, f\left(x\right)=\left(3x^2+1\right)+C, where C is an arbitrary constant.)
Once you've gotten all that out of the way and understand the previous paragraph, it should be pretty clear how the two are different. We can define \displaystyle\int_a^b\left(f\left(x\right)\right) \cdot\mathrm{d}x=F\left(b\right)-F\left(a\right), where F is any antiderivative of f. It's easy to show that we get the same value for any two antiderivatives F1 and F2, since if we define k to be the constant by which F1 and F2 differ, the k's cancel out and we're left with the same value regardless of what F1 and F2 we choose.
Another point I want to drive home is that the only reason we use 3x+C instead of, say, 3x+1+C is because the former is the most convenient to write. I sort of explained this earlier, but it's pretty important if you want to know about the nature behind the C.
