Integral of $$$a^{x} - 1$$$ with respect to $$$x$$$
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Find $$$\int \left(a^{x} - 1\right)\, dx$$$.
Solution
Integrate term by term:
$${\color{red}{\int{\left(a^{x} - 1\right)d x}}} = {\color{red}{\left(- \int{1 d x} + \int{a^{x} d x}\right)}}$$
Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:
$$\int{a^{x} d x} - {\color{red}{\int{1 d x}}} = \int{a^{x} d x} - {\color{red}{x}}$$
Apply the exponential rule $$$\int{a^{x} d x} = \frac{a^{x}}{\ln{\left(a \right)}}$$$ with $$$a=a$$$:
$$- x + {\color{red}{\int{a^{x} d x}}} = - x + {\color{red}{\frac{a^{x}}{\ln{\left(a \right)}}}}$$
Therefore,
$$\int{\left(a^{x} - 1\right)d x} = \frac{a^{x}}{\ln{\left(a \right)}} - x$$
Add the constant of integration:
$$\int{\left(a^{x} - 1\right)d x} = \frac{a^{x}}{\ln{\left(a \right)}} - x+C$$
Answer
$$$\int \left(a^{x} - 1\right)\, dx = \left(\frac{a^{x}}{\ln\left(a\right)} - x\right) + C$$$A