Integral of e^(-3x) dx

1. Let $u = - 3 x$.

   Then let $du = - 3 dx$ and substitute $- \frac{du}{3}$:

   $$\int \left(- \frac{e^{u}}{3}\right)\, du$$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\text{False}$$

      1. The integral of the exponential function is itself.

         $$\int e^{u}\, du = e^{u}$$

      So, the result is: $- \frac{e^{u}}{3}$

   Now substitute $u$ back in:

   $$- \frac{e^{- 3 x}}{3}$$

2. Add the constant of integration:

   $$- \frac{e^{- 3 x}}{3}+ \mathrm{constant}$$

The answer is:

$$- \frac{e^{- 3 x}}{3}+ \mathrm{constant}$$