Integral of $$$2^{x} - 1$$$

Find $$$\int \left(2^{x} - 1\right)\, dx$$$.

Integrate term by term:

$${\color{red}{\int{\left(2^{x} - 1\right)d x}}} = {\color{red}{\left(- \int{1 d x} + \int{2^{x} d x}\right)}}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:

$$\int{2^{x} d x} - {\color{red}{\int{1 d x}}} = \int{2^{x} d x} - {\color{red}{x}}$$

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

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

Therefore,

$$\int{\left(2^{x} - 1\right)d x} = \frac{2^{x}}{\ln{\left(2 \right)}} - x$$

Add the constant of integration:

$$\int{\left(2^{x} - 1\right)d x} = \frac{2^{x}}{\ln{\left(2 \right)}} - x+C$$

$$$\int \left(2^{x} - 1\right)\, dx = \left(\frac{2^{x}}{\ln\left(2\right)} - x\right) + C$$$A