Integrating e values with variable exponents: x = int (9 - 9e^{-3t}) dt

[TreeM]

When integrating, I'm confused as to when e^x remains as e^x, or when the exponent is brought down the front to multiply with the constant.

 

[BigBeachBanana]

When integrating, I'm confused as to when e^x remains as e^x, or when the exponent is brought down the front to multiply with the constant.View attachment 37392

It's u-subtitution.
[math]\int e^{-3t} \, dt \\ \text{Let u = -3t} \implies du = -3dt\\ \int -\dfrac{1}{3}e^{u} \, du = -\dfrac{1}{3}e^{u} + C = -\dfrac{1}{3}e^{-3t} + C \\[/math]