Integral de $$$\frac{\sqrt{2} r}{2 \left(- a + r\right)}$$$ con respecto a $$$r$$$

Halla $$$\int \frac{\sqrt{2} r}{2 \left(- a + r\right)}\, dr$$$.

Aplica la regla del factor constante $$$\int c f{\left(r \right)}\, dr = c \int f{\left(r \right)}\, dr$$$ con $$$c=\frac{\sqrt{2}}{2}$$$ y $$$f{\left(r \right)} = \frac{r}{- a + r}$$$:

$${\color{red}{\int{\frac{\sqrt{2} r}{2 \left(- a + r\right)} d r}}} = {\color{red}{\left(\frac{\sqrt{2} \int{\frac{r}{- a + r} d r}}{2}\right)}}$$

Reescribe y separa la fracción:

$$\frac{\sqrt{2} {\color{red}{\int{\frac{r}{- a + r} d r}}}}{2} = \frac{\sqrt{2} {\color{red}{\int{\left(\frac{a}{- a + r} + 1\right)d r}}}}{2}$$

Integra término a término:

$$\frac{\sqrt{2} {\color{red}{\int{\left(\frac{a}{- a + r} + 1\right)d r}}}}{2} = \frac{\sqrt{2} {\color{red}{\left(\int{1 d r} + \int{\frac{a}{- a + r} d r}\right)}}}{2}$$

Aplica la regla de la constante $$$\int c\, dr = c r$$$ con $$$c=1$$$:

$$\frac{\sqrt{2} \left(\int{\frac{a}{- a + r} d r} + {\color{red}{\int{1 d r}}}\right)}{2} = \frac{\sqrt{2} \left(\int{\frac{a}{- a + r} d r} + {\color{red}{r}}\right)}{2}$$

Aplica la regla del factor constante $$$\int c f{\left(r \right)}\, dr = c \int f{\left(r \right)}\, dr$$$ con $$$c=a$$$ y $$$f{\left(r \right)} = \frac{1}{- a + r}$$$:

$$\frac{\sqrt{2} \left(r + {\color{red}{\int{\frac{a}{- a + r} d r}}}\right)}{2} = \frac{\sqrt{2} \left(r + {\color{red}{a \int{\frac{1}{- a + r} d r}}}\right)}{2}$$

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

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

La integral puede reescribirse como

$$\frac{\sqrt{2} \left(a {\color{red}{\int{\frac{1}{- a + r} d r}}} + r\right)}{2} = \frac{\sqrt{2} \left(a {\color{red}{\int{\frac{1}{u} d u}}} + r\right)}{2}$$

La integral de $$$\frac{1}{u}$$$ es $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$\frac{\sqrt{2} \left(a {\color{red}{\int{\frac{1}{u} d u}}} + r\right)}{2} = \frac{\sqrt{2} \left(a {\color{red}{\ln{\left(\left|{u}\right| \right)}}} + r\right)}{2}$$

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

$$\frac{\sqrt{2} \left(a \ln{\left(\left|{{\color{red}{u}}}\right| \right)} + r\right)}{2} = \frac{\sqrt{2} \left(a \ln{\left(\left|{{\color{red}{\left(- a + r\right)}}}\right| \right)} + r\right)}{2}$$

Por lo tanto,

$$\int{\frac{\sqrt{2} r}{2 \left(- a + r\right)} d r} = \frac{\sqrt{2} \left(a \ln{\left(\left|{a - r}\right| \right)} + r\right)}{2}$$

Añade la constante de integración:

$$\int{\frac{\sqrt{2} r}{2 \left(- a + r\right)} d r} = \frac{\sqrt{2} \left(a \ln{\left(\left|{a - r}\right| \right)} + r\right)}{2}+C$$

$$$\int \frac{\sqrt{2} r}{2 \left(- a + r\right)}\, dr = \frac{\sqrt{2} \left(a \ln\left(\left|{a - r}\right|\right) + r\right)}{2} + C$$$A