Integral de $$$\frac{17 \sqrt{2}}{8 x \ln\left(x\right)}$$$

Encontre $$$\int \frac{17 \sqrt{2}}{8 x \ln\left(x\right)}\, dx$$$.

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{17 \sqrt{2}}{8}$$$ e $$$f{\left(x \right)} = \frac{1}{x \ln{\left(x \right)}}$$$:

$${\color{red}{\int{\frac{17 \sqrt{2}}{8 x \ln{\left(x \right)}} d x}}} = {\color{red}{\left(\frac{17 \sqrt{2} \int{\frac{1}{x \ln{\left(x \right)}} d x}}{8}\right)}}$$

Seja $$$u=\ln{\left(x \right)}$$$.

Então $$$du=\left(\ln{\left(x \right)}\right)^{\prime }dx = \frac{dx}{x}$$$ (veja os passos »), e obtemos $$$\frac{dx}{x} = du$$$.

Portanto,

$$\frac{17 \sqrt{2} {\color{red}{\int{\frac{1}{x \ln{\left(x \right)}} d x}}}}{8} = \frac{17 \sqrt{2} {\color{red}{\int{\frac{1}{u} d u}}}}{8}$$

A integral de $$$\frac{1}{u}$$$ é $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$\frac{17 \sqrt{2} {\color{red}{\int{\frac{1}{u} d u}}}}{8} = \frac{17 \sqrt{2} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{8}$$

Recorde que $$$u=\ln{\left(x \right)}$$$:

$$\frac{17 \sqrt{2} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{8} = \frac{17 \sqrt{2} \ln{\left(\left|{{\color{red}{\ln{\left(x \right)}}}}\right| \right)}}{8}$$

Portanto,

$$\int{\frac{17 \sqrt{2}}{8 x \ln{\left(x \right)}} d x} = \frac{17 \sqrt{2} \ln{\left(\left|{\ln{\left(x \right)}}\right| \right)}}{8}$$

Adicione a constante de integração:

$$\int{\frac{17 \sqrt{2}}{8 x \ln{\left(x \right)}} d x} = \frac{17 \sqrt{2} \ln{\left(\left|{\ln{\left(x \right)}}\right| \right)}}{8}+C$$

$$$\int \frac{17 \sqrt{2}}{8 x \ln\left(x\right)}\, dx = \frac{17 \sqrt{2} \ln\left(\left|{\ln\left(x\right)}\right|\right)}{8} + C$$$A