Integral of $$$2 e^{2 t - 4 u}$$$ with respect to $$$t$$$

Find $$$\int 2 e^{2 t - 4 u}\, dt$$$.

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=2$$$ and $$$f{\left(t \right)} = e^{2 t - 4 u}$$$:

$${\color{red}{\int{2 e^{2 t - 4 u} d t}}} = {\color{red}{\left(2 \int{e^{2 t - 4 u} d t}\right)}}$$

Let $$$v=2 t - 4 u$$$.

Then $$$dv=\left(2 t - 4 u\right)^{\prime }dt = 2 dt$$$ (steps can be seen »), and we have that $$$dt = \frac{dv}{2}$$$.

So,

$$2 {\color{red}{\int{e^{2 t - 4 u} d t}}} = 2 {\color{red}{\int{\frac{e^{v}}{2} d v}}}$$

Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(v \right)} = e^{v}$$$:

$$2 {\color{red}{\int{\frac{e^{v}}{2} d v}}} = 2 {\color{red}{\left(\frac{\int{e^{v} d v}}{2}\right)}}$$

The integral of the exponential function is $$$\int{e^{v} d v} = e^{v}$$$:

$${\color{red}{\int{e^{v} d v}}} = {\color{red}{e^{v}}}$$

Recall that $$$v=2 t - 4 u$$$:

$$e^{{\color{red}{v}}} = e^{{\color{red}{\left(2 t - 4 u\right)}}}$$

Therefore,

$$\int{2 e^{2 t - 4 u} d t} = e^{2 t - 4 u}$$

Add the constant of integration:

$$\int{2 e^{2 t - 4 u} d t} = e^{2 t - 4 u}+C$$

$$$\int 2 e^{2 t - 4 u}\, dt = e^{2 t - 4 u} + C$$$A