Integral de $$$- \frac{24 x}{\left(x - 5\right) \left(x - 3\right)}$$$
Calculadora relacionada: Calculadora de integrales definidas e impropias
Tu entrada
Halla $$$\int \left(- \frac{24 x}{\left(x - 5\right) \left(x - 3\right)}\right)\, dx$$$.
Solución
Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=-24$$$ y $$$f{\left(x \right)} = \frac{x}{\left(x - 5\right) \left(x - 3\right)}$$$:
$${\color{red}{\int{\left(- \frac{24 x}{\left(x - 5\right) \left(x - 3\right)}\right)d x}}} = {\color{red}{\left(- 24 \int{\frac{x}{\left(x - 5\right) \left(x - 3\right)} d x}\right)}}$$
Realizar la descomposición en fracciones parciales (los pasos pueden verse »):
$$- 24 {\color{red}{\int{\frac{x}{\left(x - 5\right) \left(x - 3\right)} d x}}} = - 24 {\color{red}{\int{\left(- \frac{3}{2 \left(x - 3\right)} + \frac{5}{2 \left(x - 5\right)}\right)d x}}}$$
Integra término a término:
$$- 24 {\color{red}{\int{\left(- \frac{3}{2 \left(x - 3\right)} + \frac{5}{2 \left(x - 5\right)}\right)d x}}} = - 24 {\color{red}{\left(\int{\frac{5}{2 \left(x - 5\right)} d x} - \int{\frac{3}{2 \left(x - 3\right)} d x}\right)}}$$
Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{3}{2}$$$ y $$$f{\left(x \right)} = \frac{1}{x - 3}$$$:
$$- 24 \int{\frac{5}{2 \left(x - 5\right)} d x} + 24 {\color{red}{\int{\frac{3}{2 \left(x - 3\right)} d x}}} = - 24 \int{\frac{5}{2 \left(x - 5\right)} d x} + 24 {\color{red}{\left(\frac{3 \int{\frac{1}{x - 3} d x}}{2}\right)}}$$
Sea $$$u=x - 3$$$.
Entonces $$$du=\left(x - 3\right)^{\prime }dx = 1 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = du$$$.
Por lo tanto,
$$- 24 \int{\frac{5}{2 \left(x - 5\right)} d x} + 36 {\color{red}{\int{\frac{1}{x - 3} d x}}} = - 24 \int{\frac{5}{2 \left(x - 5\right)} d x} + 36 {\color{red}{\int{\frac{1}{u} d u}}}$$
La integral de $$$\frac{1}{u}$$$ es $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$- 24 \int{\frac{5}{2 \left(x - 5\right)} d x} + 36 {\color{red}{\int{\frac{1}{u} d u}}} = - 24 \int{\frac{5}{2 \left(x - 5\right)} d x} + 36 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$
Recordemos que $$$u=x - 3$$$:
$$36 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} - 24 \int{\frac{5}{2 \left(x - 5\right)} d x} = 36 \ln{\left(\left|{{\color{red}{\left(x - 3\right)}}}\right| \right)} - 24 \int{\frac{5}{2 \left(x - 5\right)} d x}$$
Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{5}{2}$$$ y $$$f{\left(x \right)} = \frac{1}{x - 5}$$$:
$$36 \ln{\left(\left|{x - 3}\right| \right)} - 24 {\color{red}{\int{\frac{5}{2 \left(x - 5\right)} d x}}} = 36 \ln{\left(\left|{x - 3}\right| \right)} - 24 {\color{red}{\left(\frac{5 \int{\frac{1}{x - 5} d x}}{2}\right)}}$$
Sea $$$u=x - 5$$$.
Entonces $$$du=\left(x - 5\right)^{\prime }dx = 1 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = du$$$.
La integral se convierte en
$$36 \ln{\left(\left|{x - 3}\right| \right)} - 60 {\color{red}{\int{\frac{1}{x - 5} d x}}} = 36 \ln{\left(\left|{x - 3}\right| \right)} - 60 {\color{red}{\int{\frac{1}{u} d u}}}$$
La integral de $$$\frac{1}{u}$$$ es $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$36 \ln{\left(\left|{x - 3}\right| \right)} - 60 {\color{red}{\int{\frac{1}{u} d u}}} = 36 \ln{\left(\left|{x - 3}\right| \right)} - 60 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$
Recordemos que $$$u=x - 5$$$:
$$36 \ln{\left(\left|{x - 3}\right| \right)} - 60 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = 36 \ln{\left(\left|{x - 3}\right| \right)} - 60 \ln{\left(\left|{{\color{red}{\left(x - 5\right)}}}\right| \right)}$$
Por lo tanto,
$$\int{\left(- \frac{24 x}{\left(x - 5\right) \left(x - 3\right)}\right)d x} = - 60 \ln{\left(\left|{x - 5}\right| \right)} + 36 \ln{\left(\left|{x - 3}\right| \right)}$$
Añade la constante de integración:
$$\int{\left(- \frac{24 x}{\left(x - 5\right) \left(x - 3\right)}\right)d x} = - 60 \ln{\left(\left|{x - 5}\right| \right)} + 36 \ln{\left(\left|{x - 3}\right| \right)}+C$$
Respuesta
$$$\int \left(- \frac{24 x}{\left(x - 5\right) \left(x - 3\right)}\right)\, dx = \left(- 60 \ln\left(\left|{x - 5}\right|\right) + 36 \ln\left(\left|{x - 3}\right|\right)\right) + C$$$A