Integral de $$$5^{- x} x$$$

Halla $$$\int 5^{- x} x\, dx$$$.

Para la integral $$$\int{5^{- x} x d x}$$$, utiliza la integración por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sean $$$\operatorname{u}=x$$$ y $$$\operatorname{dv}=5^{- x} dx$$$.

Entonces $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (los pasos pueden verse ») y $$$\operatorname{v}=\int{5^{- x} d x}=- \frac{5^{- x}}{\ln{\left(5 \right)}}$$$ (los pasos pueden verse »).

Entonces,

$${\color{red}{\int{5^{- x} x d x}}}={\color{red}{\left(x \cdot \left(- \frac{5^{- x}}{\ln{\left(5 \right)}}\right)-\int{\left(- \frac{5^{- x}}{\ln{\left(5 \right)}}\right) \cdot 1 d x}\right)}}={\color{red}{\left(- \int{\left(- \frac{5^{- x}}{\ln{\left(5 \right)}}\right)d x} - \frac{5^{- x} x}{\ln{\left(5 \right)}}\right)}}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=- \frac{1}{\ln{\left(5 \right)}}$$$ y $$$f{\left(x \right)} = 5^{- x}$$$:

$$- {\color{red}{\int{\left(- \frac{5^{- x}}{\ln{\left(5 \right)}}\right)d x}}} - \frac{5^{- x} x}{\ln{\left(5 \right)}} = - {\color{red}{\left(- \frac{\int{5^{- x} d x}}{\ln{\left(5 \right)}}\right)}} - \frac{5^{- x} x}{\ln{\left(5 \right)}}$$

Sea $$$u=- x$$$.

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

Entonces,

$$\frac{{\color{red}{\int{5^{- x} d x}}}}{\ln{\left(5 \right)}} - \frac{5^{- x} x}{\ln{\left(5 \right)}} = \frac{{\color{red}{\int{\left(- 5^{u}\right)d u}}}}{\ln{\left(5 \right)}} - \frac{5^{- x} x}{\ln{\left(5 \right)}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=-1$$$ y $$$f{\left(u \right)} = 5^{u}$$$:

$$\frac{{\color{red}{\int{\left(- 5^{u}\right)d u}}}}{\ln{\left(5 \right)}} - \frac{5^{- x} x}{\ln{\left(5 \right)}} = \frac{{\color{red}{\left(- \int{5^{u} d u}\right)}}}{\ln{\left(5 \right)}} - \frac{5^{- x} x}{\ln{\left(5 \right)}}$$

Apply the exponential rule $$$\int{a^{u} d u} = \frac{a^{u}}{\ln{\left(a \right)}}$$$ with $$$a=5$$$:

$$- \frac{{\color{red}{\int{5^{u} d u}}}}{\ln{\left(5 \right)}} - \frac{5^{- x} x}{\ln{\left(5 \right)}} = - \frac{{\color{red}{\frac{5^{u}}{\ln{\left(5 \right)}}}}}{\ln{\left(5 \right)}} - \frac{5^{- x} x}{\ln{\left(5 \right)}}$$

Recordemos que $$$u=- x$$$:

$$- \frac{5^{{\color{red}{u}}}}{\ln{\left(5 \right)}^{2}} - \frac{5^{- x} x}{\ln{\left(5 \right)}} = - \frac{5^{{\color{red}{\left(- x\right)}}}}{\ln{\left(5 \right)}^{2}} - \frac{5^{- x} x}{\ln{\left(5 \right)}}$$

Por lo tanto,

$$\int{5^{- x} x d x} = - \frac{5^{- x} x}{\ln{\left(5 \right)}} - \frac{5^{- x}}{\ln{\left(5 \right)}^{2}}$$

Simplificar:

$$\int{5^{- x} x d x} = \frac{5^{- x} \left(- x \ln{\left(5 \right)} - 1\right)}{\ln{\left(5 \right)}^{2}}$$

Añade la constante de integración:

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

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