Integral de $$$\frac{4 \ln\left(x\right)}{x^{4}}$$$

Encontre $$$\int \frac{4 \ln\left(x\right)}{x^{4}}\, dx$$$.

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=4$$$ e $$$f{\left(x \right)} = \frac{\ln{\left(x \right)}}{x^{4}}$$$:

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

Para a integral $$$\int{\frac{\ln{\left(x \right)}}{x^{4}} d x}$$$, use integração por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sejam $$$\operatorname{u}=\ln{\left(x \right)}$$$ e $$$\operatorname{dv}=\frac{dx}{x^{4}}$$$.

Então $$$\operatorname{du}=\left(\ln{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x}$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{\frac{1}{x^{4}} d x}=- \frac{1}{3 x^{3}}$$$ (os passos podem ser vistos »).

Assim,

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

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=- \frac{1}{3}$$$ e $$$f{\left(x \right)} = \frac{1}{x^{4}}$$$:

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

Aplique a regra da potência $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=-4$$$:

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

Portanto,

$$\int{\frac{4 \ln{\left(x \right)}}{x^{4}} d x} = - \frac{4 \ln{\left(x \right)}}{3 x^{3}} - \frac{4}{9 x^{3}}$$

Simplifique:

$$\int{\frac{4 \ln{\left(x \right)}}{x^{4}} d x} = \frac{4 \left(- 3 \ln{\left(x \right)} - 1\right)}{9 x^{3}}$$

Adicione a constante de integração:

$$\int{\frac{4 \ln{\left(x \right)}}{x^{4}} d x} = \frac{4 \left(- 3 \ln{\left(x \right)} - 1\right)}{9 x^{3}}+C$$

$$$\int \frac{4 \ln\left(x\right)}{x^{4}}\, dx = \frac{4 \left(- 3 \ln\left(x\right) - 1\right)}{9 x^{3}} + C$$$A