Integral of $$$2^{x} - 4^{x}$$$

Find $$$\int \left(2^{x} - 4^{x}\right)\, dx$$$.

Integrate term by term:

$${\color{red}{\int{\left(2^{x} - 4^{x}\right)d x}}} = {\color{red}{\left(\int{2^{x} d x} - \int{4^{x} d x}\right)}}$$

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

$$- \int{4^{x} d x} + {\color{red}{\int{2^{x} d x}}} = - \int{4^{x} d x} + {\color{red}{\frac{2^{x}}{\ln{\left(2 \right)}}}}$$

Apply the exponential rule $$$\int{a^{x} d x} = \frac{a^{x}}{\ln{\left(a \right)}}$$$ with $$$a=4$$$:

$$\frac{2^{x}}{\ln{\left(2 \right)}} - {\color{red}{\int{4^{x} d x}}} = \frac{2^{x}}{\ln{\left(2 \right)}} - {\color{red}{\frac{4^{x}}{\ln{\left(4 \right)}}}}$$

Therefore,

$$\int{\left(2^{x} - 4^{x}\right)d x} = \frac{2^{x}}{\ln{\left(2 \right)}} - \frac{4^{x}}{\ln{\left(4 \right)}}$$

Simplify:

$$\int{\left(2^{x} - 4^{x}\right)d x} = \frac{2 \cdot 2^{x} - 4^{x}}{2 \ln{\left(2 \right)}}$$

Add the constant of integration:

$$\int{\left(2^{x} - 4^{x}\right)d x} = \frac{2 \cdot 2^{x} - 4^{x}}{2 \ln{\left(2 \right)}}+C$$

$$$\int \left(2^{x} - 4^{x}\right)\, dx = \frac{2 \cdot 2^{x} - 4^{x}}{2 \ln\left(2\right)} + C$$$A