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

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

Integrate term by term:

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

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

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

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

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

The integral becomes

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

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

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

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

$$- \int{5 x d x} + \int{2 e^{x} d x} + 10 {\color{red}{\int{e^{u} d u}}} = - \int{5 x d x} + \int{2 e^{x} d x} + 10 {\color{red}{e^{u}}}$$

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

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

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

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

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:

$$\int{2 e^{x} d x} - 5 {\color{red}{\int{x d x}}} + 10 e^{- x}=\int{2 e^{x} d x} - 5 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}} + 10 e^{- x}=\int{2 e^{x} d x} - 5 {\color{red}{\left(\frac{x^{2}}{2}\right)}} + 10 e^{- x}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=2$$$ and $$$f{\left(x \right)} = e^{x}$$$:

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

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

$$- \frac{5 x^{2}}{2} + 2 {\color{red}{\int{e^{x} d x}}} + 10 e^{- x} = - \frac{5 x^{2}}{2} + 2 {\color{red}{e^{x}}} + 10 e^{- x}$$

Therefore,

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

Add the constant of integration:

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

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