Integral of $$$e^{- \frac{141 p t}{800} + \frac{1673}{500}}$$$ with respect to $$$t$$$

Find $$$\int e^{- \frac{141 p t}{800} + \frac{1673}{500}}\, dt$$$.

Let $$$u=- \frac{141 p t}{800} + \frac{1673}{500}$$$.

Then $$$du=\left(- \frac{141 p t}{800} + \frac{1673}{500}\right)^{\prime }dt = - \frac{141 p}{800} dt$$$ (steps can be seen »), and we have that $$$dt = - \frac{800 du}{141 p}$$$.

Thus,

$${\color{red}{\int{e^{- \frac{141 p t}{800} + \frac{1673}{500}} d t}}} = {\color{red}{\int{\left(- \frac{800 e^{u}}{141 p}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=- \frac{800}{141 p}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\left(- \frac{800 e^{u}}{141 p}\right)d u}}} = {\color{red}{\left(- \frac{800 \int{e^{u} d u}}{141 p}\right)}}$$

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

$$- \frac{800 {\color{red}{\int{e^{u} d u}}}}{141 p} = - \frac{800 {\color{red}{e^{u}}}}{141 p}$$

Recall that $$$u=- \frac{141 p t}{800} + \frac{1673}{500}$$$:

$$- \frac{800 e^{{\color{red}{u}}}}{141 p} = - \frac{800 e^{{\color{red}{\left(- \frac{141 p t}{800} + \frac{1673}{500}\right)}}}}{141 p}$$

Therefore,

$$\int{e^{- \frac{141 p t}{800} + \frac{1673}{500}} d t} = - \frac{800 e^{- \frac{141 p t}{800} + \frac{1673}{500}}}{141 p}$$

Add the constant of integration:

$$\int{e^{- \frac{141 p t}{800} + \frac{1673}{500}} d t} = - \frac{800 e^{- \frac{141 p t}{800} + \frac{1673}{500}}}{141 p}+C$$

$$$\int e^{- \frac{141 p t}{800} + \frac{1673}{500}}\, dt = - \frac{800 e^{- \frac{141 p t}{800} + \frac{1673}{500}}}{141 p} + C$$$A