Integral de $$$\frac{x^{6} - 1}{x^{2} + 1}$$$

Halla $$$\int \frac{x^{6} - 1}{x^{2} + 1}\, dx$$$.

Como el grado del numerador no es menor que el grado del denominador, realiza la división larga de polinomios (los pasos pueden verse »):

$${\color{red}{\int{\frac{x^{6} - 1}{x^{2} + 1} d x}}} = {\color{red}{\int{\left(x^{4} - x^{2} + 1 - \frac{2}{x^{2} + 1}\right)d x}}}$$

Integra término a término:

$${\color{red}{\int{\left(x^{4} - x^{2} + 1 - \frac{2}{x^{2} + 1}\right)d x}}} = {\color{red}{\left(\int{1 d x} - \int{x^{2} d x} + \int{x^{4} d x} - \int{\frac{2}{x^{2} + 1} d x}\right)}}$$

Aplica la regla de la constante $$$\int c\, dx = c x$$$ con $$$c=1$$$:

$$- \int{x^{2} d x} + \int{x^{4} d x} - \int{\frac{2}{x^{2} + 1} d x} + {\color{red}{\int{1 d x}}} = - \int{x^{2} d x} + \int{x^{4} d x} - \int{\frac{2}{x^{2} + 1} d x} + {\color{red}{x}}$$

Aplica la regla de la potencia $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=4$$$:

$$x - \int{x^{2} d x} - \int{\frac{2}{x^{2} + 1} d x} + {\color{red}{\int{x^{4} d x}}}=x - \int{x^{2} d x} - \int{\frac{2}{x^{2} + 1} d x} + {\color{red}{\frac{x^{1 + 4}}{1 + 4}}}=x - \int{x^{2} d x} - \int{\frac{2}{x^{2} + 1} d x} + {\color{red}{\left(\frac{x^{5}}{5}\right)}}$$

Aplica la regla de la potencia $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=2$$$:

$$\frac{x^{5}}{5} + x - \int{\frac{2}{x^{2} + 1} d x} - {\color{red}{\int{x^{2} d x}}}=\frac{x^{5}}{5} + x - \int{\frac{2}{x^{2} + 1} d x} - {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=\frac{x^{5}}{5} + x - \int{\frac{2}{x^{2} + 1} d x} - {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

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

$$\frac{x^{5}}{5} - \frac{x^{3}}{3} + x - {\color{red}{\int{\frac{2}{x^{2} + 1} d x}}} = \frac{x^{5}}{5} - \frac{x^{3}}{3} + x - {\color{red}{\left(2 \int{\frac{1}{x^{2} + 1} d x}\right)}}$$

La integral de $$$\frac{1}{x^{2} + 1}$$$ es $$$\int{\frac{1}{x^{2} + 1} d x} = \operatorname{atan}{\left(x \right)}$$$:

$$\frac{x^{5}}{5} - \frac{x^{3}}{3} + x - 2 {\color{red}{\int{\frac{1}{x^{2} + 1} d x}}} = \frac{x^{5}}{5} - \frac{x^{3}}{3} + x - 2 {\color{red}{\operatorname{atan}{\left(x \right)}}}$$

Por lo tanto,

$$\int{\frac{x^{6} - 1}{x^{2} + 1} d x} = \frac{x^{5}}{5} - \frac{x^{3}}{3} + x - 2 \operatorname{atan}{\left(x \right)}$$

Añade la constante de integración:

$$\int{\frac{x^{6} - 1}{x^{2} + 1} d x} = \frac{x^{5}}{5} - \frac{x^{3}}{3} + x - 2 \operatorname{atan}{\left(x \right)}+C$$

$$$\int \frac{x^{6} - 1}{x^{2} + 1}\, dx = \left(\frac{x^{5}}{5} - \frac{x^{3}}{3} + x - 2 \operatorname{atan}{\left(x \right)}\right) + C$$$A