Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.3 Derivatives Of Inverse Functions; Derivatives And Integrals Involving Exponential Functions - Exercises Set 6.3 - Page 433: 63

$$ - \frac{1}{5}{e^{ - 5x}} + C$$

$$\eqalign{ & \int {{e^{ - 5x}}dx;\,\,{\text{ with }}u = - 5x} \cr & u = - 5x \to du = - 5dx,\,\,dx = - \frac{1}{5}du \cr & {\text{using the indicated substitution in the book }} \cr & \int {{e^{ - 5x}}} dx = \int {{e^u}\left( { - \frac{1}{5}du} \right)} \cr & = - \frac{1}{5}\int {{e^u}} du \cr & {\text{integrate}} \cr & = - \frac{1}{5}{e^u} + C \cr & {\text{substituting }}u = - 5x \cr & = - \frac{1}{5}{e^{ - 5x}} + C \cr} $$