Integral de $$$x y^{x}$$$ em relação a $$$x$$$

Encontre $$$\int x y^{x}\, dx$$$.

Para a integral $$$\int{x y^{x} d x}$$$, use integração por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sejam $$$\operatorname{u}=x$$$ e $$$\operatorname{dv}=y^{x} dx$$$.

Então $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{y^{x} d x}=\frac{y^{x}}{\ln{\left(y \right)}}$$$ (os passos podem ser vistos »).

Logo,

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

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{1}{\ln{\left(y \right)}}$$$ e $$$f{\left(x \right)} = y^{x}$$$:

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

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

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

Portanto,

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

Simplifique:

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

Adicione a constante de integração:

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

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