Integral of $$$2^{x} 4^{- x} - 25$$$

Find $$$\int \left(2^{x} 4^{- x} - 25\right)\, dx$$$.

The input is rewritten: $$$\int{\left(2^{x} 4^{- x} - 25\right)d x}=\int{\left(-25 + 2^{- x}\right)d x}$$$.

Integrate term by term:

$${\color{red}{\int{\left(-25 + 2^{- x}\right)d x}}} = {\color{red}{\left(- \int{25 d x} + \int{2^{- x} d x}\right)}}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=25$$$:

$$\int{2^{- x} d x} - {\color{red}{\int{25 d x}}} = \int{2^{- x} d x} - {\color{red}{\left(25 x\right)}}$$

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

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

The integral becomes

$$- 25 x + {\color{red}{\int{2^{- x} d x}}} = - 25 x + {\color{red}{\int{\left(- 2^{u}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=-1$$$ and $$$f{\left(u \right)} = 2^{u}$$$:

$$- 25 x + {\color{red}{\int{\left(- 2^{u}\right)d u}}} = - 25 x + {\color{red}{\left(- \int{2^{u} d u}\right)}}$$

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

$$- 25 x - {\color{red}{\int{2^{u} d u}}} = - 25 x - {\color{red}{\frac{2^{u}}{\ln{\left(2 \right)}}}}$$

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

$$- 25 x - \frac{2^{{\color{red}{u}}}}{\ln{\left(2 \right)}} = - 25 x - \frac{2^{{\color{red}{\left(- x\right)}}}}{\ln{\left(2 \right)}}$$

Therefore,

$$\int{\left(-25 + 2^{- x}\right)d x} = - 25 x - \frac{2^{- x}}{\ln{\left(2 \right)}}$$

Add the constant of integration:

$$\int{\left(-25 + 2^{- x}\right)d x} = - 25 x - \frac{2^{- x}}{\ln{\left(2 \right)}}+C$$

$$$\int \left(2^{x} 4^{- x} - 25\right)\, dx = \left(- 25 x - \frac{2^{- x}}{\ln\left(2\right)}\right) + C$$$A