Evaluate the following integrals. \int e^{3-4x}dx

Step 1
Given integral:
${\displaystyle \int e^{3-4x}dx}$
Substitute t=3-4x and differentiate it w.r.t "x"
${\displaystyle \frac{dt}{dx}=\frac{d}{dx}(3-4x)}$
${\displaystyle \frac{dt}{dx}=-4}$
${\displaystyle \frac{-dt}{4}=dx}$
Therefore, the given integral becomes,
${\displaystyle \frac{-1}{4}\int e^{t}dt}$
Step 2
We know that
${\displaystyle \int e^{x}dx=e^x+c}$
Where, "c" is integration constant.
${\displaystyle \Rightarrow \frac{-1}{4}{e}^t+c}$
Substitute t=3-4x in above equation, we get
${\displaystyle \Rightarrow \frac{-1}{4}{e}^{(3-4x)}+c}$
${\displaystyle \Rightarrow \frac{-e^{(3-4x)}}{4}+c}$