Integral de $$$\frac{1}{t^{3} - t}$$$

Halla $$$\int \frac{1}{t^{3} - t}\, dt$$$.

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

$${\color{red}{\int{\frac{1}{t^{3} - t} d t}}} = {\color{red}{\int{\left(\frac{1}{2 \left(t + 1\right)} + \frac{1}{2 \left(t - 1\right)} - \frac{1}{t}\right)d t}}}$$

Integra término a término:

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

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

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

Sea $$$u=t + 1$$$.

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

Por lo tanto,

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

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

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

Recordemos que $$$u=t + 1$$$:

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

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

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

Sea $$$u=t - 1$$$.

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

La integral se convierte en

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

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

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

Recordemos que $$$u=t - 1$$$:

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

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

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

Por lo tanto,

$$\int{\frac{1}{t^{3} - t} d t} = - \ln{\left(\left|{t}\right| \right)} + \frac{\ln{\left(\left|{t - 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{t + 1}\right| \right)}}{2}$$

Añade la constante de integración:

$$\int{\frac{1}{t^{3} - t} d t} = - \ln{\left(\left|{t}\right| \right)} + \frac{\ln{\left(\left|{t - 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{t + 1}\right| \right)}}{2}+C$$

$$$\int \frac{1}{t^{3} - t}\, dt = \left(- \ln\left(\left|{t}\right|\right) + \frac{\ln\left(\left|{t - 1}\right|\right)}{2} + \frac{\ln\left(\left|{t + 1}\right|\right)}{2}\right) + C$$$A