Integral de $$$\frac{4 t^{3}}{1 - t^{4}}$$$

Halla $$$\int \frac{4 t^{3}}{1 - t^{4}}\, dt$$$.

Sea $$$u=1 - t^{4}$$$.

Entonces $$$du=\left(1 - t^{4}\right)^{\prime }dt = - 4 t^{3} dt$$$ (los pasos pueden verse »), y obtenemos que $$$t^{3} dt = - \frac{du}{4}$$$.

La integral se convierte en

$${\color{red}{\int{\frac{4 t^{3}}{1 - t^{4}} d t}}} = {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}}$$

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 $$$u=1 - t^{4}$$$:

$$- \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = - \ln{\left(\left|{{\color{red}{\left(1 - t^{4}\right)}}}\right| \right)}$$

Por lo tanto,

$$\int{\frac{4 t^{3}}{1 - t^{4}} d t} = - \ln{\left(\left|{t^{4} - 1}\right| \right)}$$

Añade la constante de integración:

$$\int{\frac{4 t^{3}}{1 - t^{4}} d t} = - \ln{\left(\left|{t^{4} - 1}\right| \right)}+C$$

$$$\int \frac{4 t^{3}}{1 - t^{4}}\, dt = - \ln\left(\left|{t^{4} - 1}\right|\right) + C$$$A