Integral de $$$\left(- a + x\right)^{- p}$$$ con respecto a $$$x$$$

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

La entrada se reescribe: $$$\int{\left(- a + x\right)^{- p} d x}=\int{\left(\frac{1}{- a + x}\right)^{p} d x}$$$.

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

Entonces $$$du=\left(- a + x\right)^{\prime }dx = 1 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = du$$$.

Por lo tanto,

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

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

Entonces $$$dv=\left(\frac{1}{u}\right)^{\prime }du = - \frac{1}{u^{2}} du$$$ (los pasos pueden verse »), y obtenemos que $$$\frac{du}{u^{2}} = - dv$$$.

Entonces,

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

Aplica la regla del factor constante $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ con $$$c=-1$$$ y $$$f{\left(v \right)} = v^{p - 2}$$$:

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

Aplica la regla de la potencia $$$\int v^{n}\, dv = \frac{v^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=p - 2$$$:

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

Recordemos que $$$v=\frac{1}{u}$$$:

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

Recordemos que $$$u=- a + x$$$:

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

Por lo tanto,

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

Añade la constante de integración:

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

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