Integral de $$$\frac{3 - 4 x}{x^{2} - 48 x}$$$

Halla $$$\int \frac{3 - 4 x}{x^{2} - 48 x}\, dx$$$.

Reescribe el término lineal como $$$3 - 4 x=- 4 x\color{red}{+96-96}+3=- 4 x+96-93$$$ y descompón la expresión:

$${\color{red}{\int{\frac{3 - 4 x}{x^{2} - 48 x} d x}}} = {\color{red}{\int{\left(\frac{96 - 4 x}{x^{2} - 48 x} - \frac{93}{x^{2} - 48 x}\right)d x}}}$$

Integra término a término:

$${\color{red}{\int{\left(\frac{96 - 4 x}{x^{2} - 48 x} - \frac{93}{x^{2} - 48 x}\right)d x}}} = {\color{red}{\left(\int{\frac{96 - 4 x}{x^{2} - 48 x} d x} + \int{\left(- \frac{93}{x^{2} - 48 x}\right)d x}\right)}}$$

Sea $$$u=x^{2} - 48 x$$$.

Entonces $$$du=\left(x^{2} - 48 x\right)^{\prime }dx = \left(2 x - 48\right) dx$$$ (los pasos pueden verse »), y obtenemos que $$$\left(2 x - 48\right) dx = du$$$.

La integral puede reescribirse como

$$\int{\left(- \frac{93}{x^{2} - 48 x}\right)d x} + {\color{red}{\int{\frac{96 - 4 x}{x^{2} - 48 x} d x}}} = \int{\left(- \frac{93}{x^{2} - 48 x}\right)d x} + {\color{red}{\int{\left(- \frac{2}{u}\right)d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=-2$$$ y $$$f{\left(u \right)} = \frac{1}{u}$$$:

$$\int{\left(- \frac{93}{x^{2} - 48 x}\right)d x} + {\color{red}{\int{\left(- \frac{2}{u}\right)d u}}} = \int{\left(- \frac{93}{x^{2} - 48 x}\right)d x} + {\color{red}{\left(- 2 \int{\frac{1}{u} d u}\right)}}$$

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

$$\int{\left(- \frac{93}{x^{2} - 48 x}\right)d x} - 2 {\color{red}{\int{\frac{1}{u} d u}}} = \int{\left(- \frac{93}{x^{2} - 48 x}\right)d x} - 2 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

Recordemos que $$$u=x^{2} - 48 x$$$:

$$- 2 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} + \int{\left(- \frac{93}{x^{2} - 48 x}\right)d x} = - 2 \ln{\left(\left|{{\color{red}{\left(x^{2} - 48 x\right)}}}\right| \right)} + \int{\left(- \frac{93}{x^{2} - 48 x}\right)d x}$$

Simplificar el integrando:

$$- 2 \ln{\left(\left|{x^{2} - 48 x}\right| \right)} + {\color{red}{\int{\left(- \frac{93}{x^{2} - 48 x}\right)d x}}} = - 2 \ln{\left(\left|{x^{2} - 48 x}\right| \right)} + {\color{red}{\int{\left(- \frac{93}{x \left(x - 48\right)}\right)d x}}}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=-93$$$ y $$$f{\left(x \right)} = \frac{1}{x \left(x - 48\right)}$$$:

$$- 2 \ln{\left(\left|{x^{2} - 48 x}\right| \right)} + {\color{red}{\int{\left(- \frac{93}{x \left(x - 48\right)}\right)d x}}} = - 2 \ln{\left(\left|{x^{2} - 48 x}\right| \right)} + {\color{red}{\left(- 93 \int{\frac{1}{x \left(x - 48\right)} d x}\right)}}$$

Realizar la descomposición en fracciones parciales (los pasos pueden verse »):

$$- 2 \ln{\left(\left|{x^{2} - 48 x}\right| \right)} - 93 {\color{red}{\int{\frac{1}{x \left(x - 48\right)} d x}}} = - 2 \ln{\left(\left|{x^{2} - 48 x}\right| \right)} - 93 {\color{red}{\int{\left(\frac{1}{48 \left(x - 48\right)} - \frac{1}{48 x}\right)d x}}}$$

Integra término a término:

$$- 2 \ln{\left(\left|{x^{2} - 48 x}\right| \right)} - 93 {\color{red}{\int{\left(\frac{1}{48 \left(x - 48\right)} - \frac{1}{48 x}\right)d x}}} = - 2 \ln{\left(\left|{x^{2} - 48 x}\right| \right)} - 93 {\color{red}{\left(- \int{\frac{1}{48 x} d x} + \int{\frac{1}{48 \left(x - 48\right)} d x}\right)}}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{1}{48}$$$ y $$$f{\left(x \right)} = \frac{1}{x}$$$:

$$- 2 \ln{\left(\left|{x^{2} - 48 x}\right| \right)} - 93 \int{\frac{1}{48 \left(x - 48\right)} d x} + 93 {\color{red}{\int{\frac{1}{48 x} d x}}} = - 2 \ln{\left(\left|{x^{2} - 48 x}\right| \right)} - 93 \int{\frac{1}{48 \left(x - 48\right)} d x} + 93 {\color{red}{\left(\frac{\int{\frac{1}{x} d x}}{48}\right)}}$$

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

$$- 2 \ln{\left(\left|{x^{2} - 48 x}\right| \right)} - 93 \int{\frac{1}{48 \left(x - 48\right)} d x} + \frac{31 {\color{red}{\int{\frac{1}{x} d x}}}}{16} = - 2 \ln{\left(\left|{x^{2} - 48 x}\right| \right)} - 93 \int{\frac{1}{48 \left(x - 48\right)} d x} + \frac{31 {\color{red}{\ln{\left(\left|{x}\right| \right)}}}}{16}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{1}{48}$$$ y $$$f{\left(x \right)} = \frac{1}{x - 48}$$$:

$$\frac{31 \ln{\left(\left|{x}\right| \right)}}{16} - 2 \ln{\left(\left|{x^{2} - 48 x}\right| \right)} - 93 {\color{red}{\int{\frac{1}{48 \left(x - 48\right)} d x}}} = \frac{31 \ln{\left(\left|{x}\right| \right)}}{16} - 2 \ln{\left(\left|{x^{2} - 48 x}\right| \right)} - 93 {\color{red}{\left(\frac{\int{\frac{1}{x - 48} d x}}{48}\right)}}$$

Sea $$$u=x - 48$$$.

Entonces $$$du=\left(x - 48\right)^{\prime }dx = 1 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = du$$$.

Entonces,

$$\frac{31 \ln{\left(\left|{x}\right| \right)}}{16} - 2 \ln{\left(\left|{x^{2} - 48 x}\right| \right)} - \frac{31 {\color{red}{\int{\frac{1}{x - 48} d x}}}}{16} = \frac{31 \ln{\left(\left|{x}\right| \right)}}{16} - 2 \ln{\left(\left|{x^{2} - 48 x}\right| \right)} - \frac{31 {\color{red}{\int{\frac{1}{u} d u}}}}{16}$$

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

$$\frac{31 \ln{\left(\left|{x}\right| \right)}}{16} - 2 \ln{\left(\left|{x^{2} - 48 x}\right| \right)} - \frac{31 {\color{red}{\int{\frac{1}{u} d u}}}}{16} = \frac{31 \ln{\left(\left|{x}\right| \right)}}{16} - 2 \ln{\left(\left|{x^{2} - 48 x}\right| \right)} - \frac{31 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{16}$$

Recordemos que $$$u=x - 48$$$:

$$\frac{31 \ln{\left(\left|{x}\right| \right)}}{16} - 2 \ln{\left(\left|{x^{2} - 48 x}\right| \right)} - \frac{31 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{16} = \frac{31 \ln{\left(\left|{x}\right| \right)}}{16} - 2 \ln{\left(\left|{x^{2} - 48 x}\right| \right)} - \frac{31 \ln{\left(\left|{{\color{red}{\left(x - 48\right)}}}\right| \right)}}{16}$$

Por lo tanto,

$$\int{\frac{3 - 4 x}{x^{2} - 48 x} d x} = \frac{31 \ln{\left(\left|{x}\right| \right)}}{16} - \frac{31 \ln{\left(\left|{x - 48}\right| \right)}}{16} - 2 \ln{\left(\left|{x^{2} - 48 x}\right| \right)}$$

Simplificar:

$$\int{\frac{3 - 4 x}{x^{2} - 48 x} d x} = \frac{31 \ln{\left(\left|{x}\right| \right)}}{16} - 2 \ln{\left(\left|{x \left(x - 48\right)}\right| \right)} - \frac{31 \ln{\left(\left|{x - 48}\right| \right)}}{16}$$

Añade la constante de integración:

$$\int{\frac{3 - 4 x}{x^{2} - 48 x} d x} = \frac{31 \ln{\left(\left|{x}\right| \right)}}{16} - 2 \ln{\left(\left|{x \left(x - 48\right)}\right| \right)} - \frac{31 \ln{\left(\left|{x - 48}\right| \right)}}{16}+C$$

$$$\int \frac{3 - 4 x}{x^{2} - 48 x}\, dx = \left(\frac{31 \ln\left(\left|{x}\right|\right)}{16} - 2 \ln\left(\left|{x \left(x - 48\right)}\right|\right) - \frac{31 \ln\left(\left|{x - 48}\right|\right)}{16}\right) + C$$$A