Integral of $$$9^{x} + 1$$$

Find $$$\int \left(9^{x} + 1\right)\, dx$$$.

Integrate term by term:

$${\color{red}{\int{\left(9^{x} + 1\right)d x}}} = {\color{red}{\left(\int{1 d x} + \int{9^{x} d x}\right)}}$$

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

$$\int{9^{x} d x} + {\color{red}{\int{1 d x}}} = \int{9^{x} d x} + {\color{red}{x}}$$

Apply the exponential rule $$$\int{a^{x} d x} = \frac{a^{x}}{\ln{\left(a \right)}}$$$ with $$$a=9$$$:

$$x + {\color{red}{\int{9^{x} d x}}} = x + {\color{red}{\frac{9^{x}}{\ln{\left(9 \right)}}}}$$

Therefore,

$$\int{\left(9^{x} + 1\right)d x} = \frac{9^{x}}{\ln{\left(9 \right)}} + x$$

Simplify:

$$\int{\left(9^{x} + 1\right)d x} = \frac{9^{x}}{2 \ln{\left(3 \right)}} + x$$

Add the constant of integration:

$$\int{\left(9^{x} + 1\right)d x} = \frac{9^{x}}{2 \ln{\left(3 \right)}} + x+C$$

$$$\int \left(9^{x} + 1\right)\, dx = \left(\frac{9^{x}}{2 \ln\left(3\right)} + x\right) + C$$$A