Integraal van $$$- x + \left(e^{x} - 1\right) e^{- x}$$$

Bepaal $$$\int \left(- x + \left(e^{x} - 1\right) e^{- x}\right)\, dx$$$.

Integreer termgewijs:

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

Pas de machtsregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=1$$$:

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

Expand the expression:

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

Integreer termgewijs:

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

Pas de constantenregel $$$\int c\, dx = c x$$$ toe met $$$c=1$$$:

$$- \frac{x^{2}}{2} - \int{e^{- x} d x} + {\color{red}{\int{1 d x}}} = - \frac{x^{2}}{2} - \int{e^{- x} d x} + {\color{red}{x}}$$

Zij $$$u=- x$$$.

Dan $$$du=\left(- x\right)^{\prime }dx = - dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = - du$$$.

Dus,

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

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=-1$$$ en $$$f{\left(u \right)} = e^{u}$$$:

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

De integraal van de exponentiële functie is $$$\int{e^{u} d u} = e^{u}$$$:

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

We herinneren eraan dat $$$u=- x$$$:

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

Dus,

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

Voeg de integratieconstante toe:

$$\int{\left(- x + \left(e^{x} - 1\right) e^{- x}\right)d x} = - \frac{x^{2}}{2} + x + e^{- x}+C$$

$$$\int \left(- x + \left(e^{x} - 1\right) e^{- x}\right)\, dx = \left(- \frac{x^{2}}{2} + x + e^{- x}\right) + C$$$A