Integral of $$$a^{- x}$$$ with respect to $$$x$$$

Find $$$\int a^{- x}\, dx$$$.

Let $$$u=- x$$$.

Then $$$du=\left(- x\right)^{\prime }dx = - dx$$$ (steps can be seen »), and we have that $$$dx = - du$$$.

The integral becomes

$${\color{red}{\int{a^{- x} d x}}} = {\color{red}{\int{\left(- a^{u}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=-1$$$ and $$$f{\left(u \right)} = a^{u}$$$:

$${\color{red}{\int{\left(- a^{u}\right)d u}}} = {\color{red}{\left(- \int{a^{u} d u}\right)}}$$

Apply the exponential rule $$$\int{a^{u} d u} = \frac{a^{u}}{\ln{\left(a \right)}}$$$ with $$$a=a$$$:

$$- {\color{red}{\int{a^{u} d u}}} = - {\color{red}{\frac{a^{u}}{\ln{\left(a \right)}}}}$$

Recall that $$$u=- x$$$:

$$- \frac{a^{{\color{red}{u}}}}{\ln{\left(a \right)}} = - \frac{a^{{\color{red}{\left(- x\right)}}}}{\ln{\left(a \right)}}$$

Therefore,

$$\int{a^{- x} d x} = - \frac{a^{- x}}{\ln{\left(a \right)}}$$

Add the constant of integration:

$$\int{a^{- x} d x} = - \frac{a^{- x}}{\ln{\left(a \right)}}+C$$

$$$\int a^{- x}\, dx = - \frac{a^{- x}}{\ln\left(a\right)} + C$$$A