Integral of $$$e^{4 x} + 5 e^{- x}$$$

Find $$$\int \left(e^{4 x} + 5 e^{- x}\right)\, dx$$$.

Integrate term by term:

$${\color{red}{\int{\left(e^{4 x} + 5 e^{- x}\right)d x}}} = {\color{red}{\left(\int{5 e^{- x} d x} + \int{e^{4 x} d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=5$$$ and $$$f{\left(x \right)} = e^{- x}$$$:

$$\int{e^{4 x} d x} + {\color{red}{\int{5 e^{- x} d x}}} = \int{e^{4 x} d x} + {\color{red}{\left(5 \int{e^{- x} d x}\right)}}$$

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

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

The integral becomes

$$\int{e^{4 x} d x} + 5 {\color{red}{\int{e^{- x} d x}}} = \int{e^{4 x} d x} + 5 {\color{red}{\int{\left(- e^{u}\right)d u}}}$$

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

$$\int{e^{4 x} d x} + 5 {\color{red}{\int{\left(- e^{u}\right)d u}}} = \int{e^{4 x} d x} + 5 {\color{red}{\left(- \int{e^{u} d u}\right)}}$$

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

$$\int{e^{4 x} d x} - 5 {\color{red}{\int{e^{u} d u}}} = \int{e^{4 x} d x} - 5 {\color{red}{e^{u}}}$$

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

$$\int{e^{4 x} d x} - 5 e^{{\color{red}{u}}} = \int{e^{4 x} d x} - 5 e^{{\color{red}{\left(- x\right)}}}$$

Let $$$u=4 x$$$.

Then $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{4}$$$.

The integral can be rewritten as

$${\color{red}{\int{e^{4 x} d x}}} - 5 e^{- x} = {\color{red}{\int{\frac{e^{u}}{4} d u}}} - 5 e^{- x}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\frac{e^{u}}{4} d u}}} - 5 e^{- x} = {\color{red}{\left(\frac{\int{e^{u} d u}}{4}\right)}} - 5 e^{- x}$$

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

$$\frac{{\color{red}{\int{e^{u} d u}}}}{4} - 5 e^{- x} = \frac{{\color{red}{e^{u}}}}{4} - 5 e^{- x}$$

Recall that $$$u=4 x$$$:

$$\frac{e^{{\color{red}{u}}}}{4} - 5 e^{- x} = \frac{e^{{\color{red}{\left(4 x\right)}}}}{4} - 5 e^{- x}$$

Therefore,

$$\int{\left(e^{4 x} + 5 e^{- x}\right)d x} = \frac{e^{4 x}}{4} - 5 e^{- x}$$

Simplify:

$$\int{\left(e^{4 x} + 5 e^{- x}\right)d x} = \frac{\left(e^{5 x} - 20\right) e^{- x}}{4}$$

Add the constant of integration:

$$\int{\left(e^{4 x} + 5 e^{- x}\right)d x} = \frac{\left(e^{5 x} - 20\right) e^{- x}}{4}+C$$

$$$\int \left(e^{4 x} + 5 e^{- x}\right)\, dx = \frac{\left(e^{5 x} - 20\right) e^{- x}}{4} + C$$$A