Integral de $$$x \ln\left(x + 1\right)$$$

Halla $$$\int x \ln\left(x + 1\right)\, dx$$$.

Para la integral $$$\int{x \ln{\left(x + 1 \right)} d x}$$$, utiliza la integración por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sean $$$\operatorname{u}=\ln{\left(x + 1 \right)}$$$ y $$$\operatorname{dv}=x dx$$$.

Entonces $$$\operatorname{du}=\left(\ln{\left(x + 1 \right)}\right)^{\prime }dx=\frac{dx}{x + 1}$$$ (los pasos pueden verse ») y $$$\operatorname{v}=\int{x d x}=\frac{x^{2}}{2}$$$ (los pasos pueden verse »).

La integral puede reescribirse como

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

Simplificar el integrando:

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

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

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

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

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

Integra término a término:

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

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

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

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

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

Sea $$$u=x + 1$$$.

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

Entonces,

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

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

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

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

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

Por lo tanto,

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

Añade la constante de integración:

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

$$$\int x \ln\left(x + 1\right)\, dx = \left(\frac{x^{2} \ln\left(x + 1\right)}{2} - \frac{x^{2}}{4} + \frac{x}{2} - \frac{\ln\left(\left|{x + 1}\right|\right)}{2}\right) + C$$$A