Integral of $$$2^{x} - 5^{x} - 10^{- x}$$$

Find $$$\int \left(2^{x} - 5^{x} - 10^{- x}\right)\, dx$$$.

Integrate term by term:

$${\color{red}{\int{\left(2^{x} - 5^{x} - 10^{- x}\right)d x}}} = {\color{red}{\left(- \int{10^{- x} d x} + \int{2^{x} d x} - \int{5^{x} d x}\right)}}$$

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

$$- \int{10^{- x} d x} - \int{5^{x} d x} + {\color{red}{\int{2^{x} d x}}} = - \int{10^{- x} d x} - \int{5^{x} d x} + {\color{red}{\frac{2^{x}}{\ln{\left(2 \right)}}}}$$

Apply the exponential rule $$$\int{a^{x} d x} = \frac{a^{x}}{\ln{\left(a \right)}}$$$ with $$$a=5$$$:

$$\frac{2^{x}}{\ln{\left(2 \right)}} - \int{10^{- x} d x} - {\color{red}{\int{5^{x} d x}}} = \frac{2^{x}}{\ln{\left(2 \right)}} - \int{10^{- x} d x} - {\color{red}{\frac{5^{x}}{\ln{\left(5 \right)}}}}$$

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

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

Therefore,

$$\frac{2^{x}}{\ln{\left(2 \right)}} - \frac{5^{x}}{\ln{\left(5 \right)}} - {\color{red}{\int{10^{- x} d x}}} = \frac{2^{x}}{\ln{\left(2 \right)}} - \frac{5^{x}}{\ln{\left(5 \right)}} - {\color{red}{\int{\left(- 10^{u}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=-1$$$ and $$$f{\left(u \right)} = 10^{u}$$$:

$$\frac{2^{x}}{\ln{\left(2 \right)}} - \frac{5^{x}}{\ln{\left(5 \right)}} - {\color{red}{\int{\left(- 10^{u}\right)d u}}} = \frac{2^{x}}{\ln{\left(2 \right)}} - \frac{5^{x}}{\ln{\left(5 \right)}} - {\color{red}{\left(- \int{10^{u} d u}\right)}}$$

Apply the exponential rule $$$\int{a^{u} d u} = \frac{a^{u}}{\ln{\left(a \right)}}$$$ with $$$a=10$$$:

$$\frac{2^{x}}{\ln{\left(2 \right)}} - \frac{5^{x}}{\ln{\left(5 \right)}} + {\color{red}{\int{10^{u} d u}}} = \frac{2^{x}}{\ln{\left(2 \right)}} - \frac{5^{x}}{\ln{\left(5 \right)}} + {\color{red}{\frac{10^{u}}{\ln{\left(10 \right)}}}}$$

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

$$\frac{2^{x}}{\ln{\left(2 \right)}} - \frac{5^{x}}{\ln{\left(5 \right)}} + \frac{10^{{\color{red}{u}}}}{\ln{\left(10 \right)}} = \frac{2^{x}}{\ln{\left(2 \right)}} - \frac{5^{x}}{\ln{\left(5 \right)}} + \frac{10^{{\color{red}{\left(- x\right)}}}}{\ln{\left(10 \right)}}$$

Therefore,

$$\int{\left(2^{x} - 5^{x} - 10^{- x}\right)d x} = \frac{2^{x}}{\ln{\left(2 \right)}} - \frac{5^{x}}{\ln{\left(5 \right)}} + \frac{10^{- x}}{\ln{\left(10 \right)}}$$

Simplify:

$$\int{\left(2^{x} - 5^{x} - 10^{- x}\right)d x} = \frac{10^{- x} 20^{x}}{\ln{\left(2 \right)}} - \frac{10^{- x} 50^{x}}{\ln{\left(5 \right)}} + \frac{10^{- x}}{\ln{\left(10 \right)}}$$

Add the constant of integration:

$$\int{\left(2^{x} - 5^{x} - 10^{- x}\right)d x} = \frac{10^{- x} 20^{x}}{\ln{\left(2 \right)}} - \frac{10^{- x} 50^{x}}{\ln{\left(5 \right)}} + \frac{10^{- x}}{\ln{\left(10 \right)}}+C$$

$$$\int \left(2^{x} - 5^{x} - 10^{- x}\right)\, dx = \left(\frac{10^{- x} 20^{x}}{\ln\left(2\right)} - \frac{10^{- x} 50^{x}}{\ln\left(5\right)} + \frac{10^{- x}}{\ln\left(10\right)}\right) + C$$$A