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

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

Integrate term by term:

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

Let $$$u=- 2 x$$$.

Then $$$du=\left(- 2 x\right)^{\prime }dx = - 2 dx$$$ (steps can be seen »), and we have that $$$dx = - \frac{du}{2}$$$.

Therefore,

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

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

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

The integral of the exponential function is $$$\int{e^{u} d u} = e^{u}$$$:

$$\int{e^{x} d x} + \frac{{\color{red}{\int{e^{u} d u}}}}{2} = \int{e^{x} d x} + \frac{{\color{red}{e^{u}}}}{2}$$

Recall that $$$u=- 2 x$$$:

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

The integral of the exponential function is $$$\int{e^{x} d x} = e^{x}$$$:

$${\color{red}{\int{e^{x} d x}}} + \frac{e^{- 2 x}}{2} = {\color{red}{e^{x}}} + \frac{e^{- 2 x}}{2}$$

Therefore,

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

Add the constant of integration:

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

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