Integraal van $$$\left(- a + x\right)^{- p}$$$ met betrekking tot $$$x$$$

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

De invoer is herschreven: $$$\int{\left(- a + x\right)^{- p} d x}=\int{\left(\frac{1}{- a + x}\right)^{p} d x}$$$.

Zij $$$u=- a + x$$$.

Dan $$$du=\left(- a + x\right)^{\prime }dx = 1 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = du$$$.

Dus,

$${\color{red}{\int{\left(\frac{1}{- a + x}\right)^{p} d x}}} = {\color{red}{\int{\left(\frac{1}{u}\right)^{p} d u}}}$$

Zij $$$v=\frac{1}{u}$$$.

Dan $$$dv=\left(\frac{1}{u}\right)^{\prime }du = - \frac{1}{u^{2}} du$$$ (de stappen zijn te zien »), en dan geldt dat $$$\frac{du}{u^{2}} = - dv$$$.

De integraal wordt

$${\color{red}{\int{\left(\frac{1}{u}\right)^{p} d u}}} = {\color{red}{\int{\left(- v^{p - 2}\right)d v}}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ toe met $$$c=-1$$$ en $$$f{\left(v \right)} = v^{p - 2}$$$:

$${\color{red}{\int{\left(- v^{p - 2}\right)d v}}} = {\color{red}{\left(- \int{v^{p - 2} d v}\right)}}$$

Pas de machtsregel $$$\int v^{n}\, dv = \frac{v^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=p - 2$$$:

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

We herinneren eraan dat $$$v=\frac{1}{u}$$$:

$$- \frac{{\color{red}{v}}^{p - 1}}{p - 1} = - \frac{{\color{red}{\frac{1}{u}}}^{p - 1}}{p - 1}$$

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

$$- \frac{\left({\color{red}{u}}^{-1}\right)^{p - 1}}{p - 1} = - \frac{\left({\color{red}{\left(- a + x\right)}}^{-1}\right)^{p - 1}}{p - 1}$$

Dus,

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

Voeg de integratieconstante toe:

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

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