Integral de $$$\frac{1}{- c + c_{max}}$$$ con respecto a $$$c$$$

Halla $$$\int \frac{1}{- c + c_{max}}\, dc$$$.

Sea $$$u=- c + c_{max}$$$.

Entonces $$$du=\left(- c + c_{max}\right)^{\prime }dc = - dc$$$ (los pasos pueden verse »), y obtenemos que $$$dc = - du$$$.

Entonces,

$${\color{red}{\int{\frac{1}{- c + c_{max}} d c}}} = {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}}$$

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

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

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

$$- {\color{red}{\int{\frac{1}{u} d u}}} = - {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

Recordemos que $$$u=- c + c_{max}$$$:

$$- \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = - \ln{\left(\left|{{\color{red}{\left(- c + c_{max}\right)}}}\right| \right)}$$

Por lo tanto,

$$\int{\frac{1}{- c + c_{max}} d c} = - \ln{\left(\left|{c - c_{max}}\right| \right)}$$

Añade la constante de integración:

$$\int{\frac{1}{- c + c_{max}} d c} = - \ln{\left(\left|{c - c_{max}}\right| \right)}+C$$

$$$\int \frac{1}{- c + c_{max}}\, dc = - \ln\left(\left|{c - c_{max}}\right|\right) + C$$$A