Integral of $$$2^{x - 3}$$$

Find $$$\int 2^{x - 3}\, dx$$$.

Let $$$u=x - 3$$$.

Then $$$du=\left(x - 3\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

Therefore,

$${\color{red}{\int{2^{x - 3} d x}}} = {\color{red}{\int{2^{u} d u}}}$$

Apply the exponential rule $$$\int{a^{u} d u} = \frac{a^{u}}{\ln{\left(a \right)}}$$$ with $$$a=2$$$:

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

Recall that $$$u=x - 3$$$:

$$\frac{2^{{\color{red}{u}}}}{\ln{\left(2 \right)}} = \frac{2^{{\color{red}{\left(x - 3\right)}}}}{\ln{\left(2 \right)}}$$

Therefore,

$$\int{2^{x - 3} d x} = \frac{2^{x - 3}}{\ln{\left(2 \right)}}$$

Add the constant of integration:

$$\int{2^{x - 3} d x} = \frac{2^{x - 3}}{\ln{\left(2 \right)}}+C$$

$$$\int 2^{x - 3}\, dx = \frac{2^{x - 3}}{\ln\left(2\right)} + C$$$A