Integral de $$$- \frac{3 x^{2}}{4} + \ln\left(x\right)$$$

Halla $$$\int \left(- \frac{3 x^{2}}{4} + \ln\left(x\right)\right)\, dx$$$.

Integra término a término:

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

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{3}{4}$$$ y $$$f{\left(x \right)} = x^{2}$$$:

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

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

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

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

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

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

Entonces,

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

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

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

Por lo tanto,

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

Simplificar:

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

Añade la constante de integración:

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

$$$\int \left(- \frac{3 x^{2}}{4} + \ln\left(x\right)\right)\, dx = x \left(- \frac{x^{2}}{4} + \ln\left(x\right) - 1\right) + C$$$A