Integral de $$$\ln\left(4 x\right)$$$

Halla $$$\int \ln\left(4 x\right)\, dx$$$.

Sea $$$u=4 x$$$.

Entonces $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{4}$$$.

Por lo tanto,

$${\color{red}{\int{\ln{\left(4 x \right)} d x}}} = {\color{red}{\int{\frac{\ln{\left(u \right)}}{4} d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{4}$$$ y $$$f{\left(u \right)} = \ln{\left(u \right)}$$$:

$${\color{red}{\int{\frac{\ln{\left(u \right)}}{4} d u}}} = {\color{red}{\left(\frac{\int{\ln{\left(u \right)} d u}}{4}\right)}}$$

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

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

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

La integral se convierte en

$$\frac{{\color{red}{\int{\ln{\left(u \right)} d u}}}}{4}=\frac{{\color{red}{\left(\ln{\left(u \right)} \cdot u-\int{u \cdot \frac{1}{u} d u}\right)}}}{4}=\frac{{\color{red}{\left(u \ln{\left(u \right)} - \int{1 d u}\right)}}}{4}$$

Aplica la regla de la constante $$$\int c\, du = c u$$$ con $$$c=1$$$:

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

Recordemos que $$$u=4 x$$$:

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

Por lo tanto,

$$\int{\ln{\left(4 x \right)} d x} = x \ln{\left(4 x \right)} - x$$

Simplificar:

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

Añade la constante de integración:

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

$$$\int \ln\left(4 x\right)\, dx = x \left(\ln\left(x\right) - 1 + 2 \ln\left(2\right)\right) + C$$$A