\(\displaystyle\int e^{-x}\,\mathrm dx\)
You probably know that the antiderivative of \(e^x\) is simply \(e^x\). So to get something that resembles this form, let \(u=-x\). Then \(\mathrm du=-\mathrm dx\), or \(-\mathrm du=\mathrm dx\).
Now the integral is
\(\displaystyle\int e^{-x}\,\mathrm dx=\int e^u(-\mathrm du)=-\int e^u\,\mathrm du=-e^u+C=-e^{-x}+C\)
