Integral de $$$\frac{x^{3}}{x + 2}$$$

Halla $$$\int \frac{x^{3}}{x + 2}\, dx$$$.

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

$${\color{red}{\int{\frac{x^{3}}{x + 2} d x}}} = {\color{red}{\int{\left(x^{2} - 2 x + 4 - \frac{8}{x + 2}\right)d x}}}$$

Integra término a término:

$${\color{red}{\int{\left(x^{2} - 2 x + 4 - \frac{8}{x + 2}\right)d x}}} = {\color{red}{\left(\int{4 d x} - \int{2 x d x} + \int{x^{2} d x} - \int{\frac{8}{x + 2} d x}\right)}}$$

Aplica la regla de la constante $$$\int c\, dx = c x$$$ con $$$c=4$$$:

$$- \int{2 x d x} + \int{x^{2} d x} - \int{\frac{8}{x + 2} d x} + {\color{red}{\int{4 d x}}} = - \int{2 x d x} + \int{x^{2} d x} - \int{\frac{8}{x + 2} d x} + {\color{red}{\left(4 x\right)}}$$

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

$$4 x - \int{2 x d x} - \int{\frac{8}{x + 2} d x} + {\color{red}{\int{x^{2} d x}}}=4 x - \int{2 x d x} - \int{\frac{8}{x + 2} d x} + {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=4 x - \int{2 x d x} - \int{\frac{8}{x + 2} d x} + {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=8$$$ y $$$f{\left(x \right)} = \frac{1}{x + 2}$$$:

$$\frac{x^{3}}{3} + 4 x - \int{2 x d x} - {\color{red}{\int{\frac{8}{x + 2} d x}}} = \frac{x^{3}}{3} + 4 x - \int{2 x d x} - {\color{red}{\left(8 \int{\frac{1}{x + 2} d x}\right)}}$$

Sea $$$u=x + 2$$$.

Entonces $$$du=\left(x + 2\right)^{\prime }dx = 1 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = du$$$.

Por lo tanto,

$$\frac{x^{3}}{3} + 4 x - \int{2 x d x} - 8 {\color{red}{\int{\frac{1}{x + 2} d x}}} = \frac{x^{3}}{3} + 4 x - \int{2 x d x} - 8 {\color{red}{\int{\frac{1}{u} d u}}}$$

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

$$\frac{x^{3}}{3} + 4 x - \int{2 x d x} - 8 {\color{red}{\int{\frac{1}{u} d u}}} = \frac{x^{3}}{3} + 4 x - \int{2 x d x} - 8 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

Recordemos que $$$u=x + 2$$$:

$$\frac{x^{3}}{3} + 4 x - 8 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} - \int{2 x d x} = \frac{x^{3}}{3} + 4 x - 8 \ln{\left(\left|{{\color{red}{\left(x + 2\right)}}}\right| \right)} - \int{2 x d x}$$

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

$$\frac{x^{3}}{3} + 4 x - 8 \ln{\left(\left|{x + 2}\right| \right)} - {\color{red}{\int{2 x d x}}} = \frac{x^{3}}{3} + 4 x - 8 \ln{\left(\left|{x + 2}\right| \right)} - {\color{red}{\left(2 \int{x d x}\right)}}$$

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

$$\frac{x^{3}}{3} + 4 x - 8 \ln{\left(\left|{x + 2}\right| \right)} - 2 {\color{red}{\int{x d x}}}=\frac{x^{3}}{3} + 4 x - 8 \ln{\left(\left|{x + 2}\right| \right)} - 2 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=\frac{x^{3}}{3} + 4 x - 8 \ln{\left(\left|{x + 2}\right| \right)} - 2 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

Por lo tanto,

$$\int{\frac{x^{3}}{x + 2} d x} = \frac{x^{3}}{3} - x^{2} + 4 x - 8 \ln{\left(\left|{x + 2}\right| \right)}$$

Añade la constante de integración:

$$\int{\frac{x^{3}}{x + 2} d x} = \frac{x^{3}}{3} - x^{2} + 4 x - 8 \ln{\left(\left|{x + 2}\right| \right)}+C$$

$$$\int \frac{x^{3}}{x + 2}\, dx = \left(\frac{x^{3}}{3} - x^{2} + 4 x - 8 \ln\left(\left|{x + 2}\right|\right)\right) + C$$$A