Integral dari $$$\frac{1}{1 - x} + \frac{1}{x \ln\left(x\right)}$$$

Temukan $$$\int \left(\frac{1}{1 - x} + \frac{1}{x \ln\left(x\right)}\right)\, dx$$$.

Integralkan suku demi suku:

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

Misalkan $$$u=1 - x$$$.

Kemudian $$$du=\left(1 - x\right)^{\prime }dx = - dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dx = - du$$$.

Integralnya menjadi

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

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=-1$$$ dan $$$f{\left(u \right)} = \frac{1}{u}$$$:

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

Integral dari $$$\frac{1}{u}$$$ adalah $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Ingat bahwa $$$u=1 - x$$$:

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

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

Kemudian $$$du=\left(\ln{\left(x \right)}\right)^{\prime }dx = \frac{dx}{x}$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$\frac{dx}{x} = du$$$.

Dengan demikian,

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

Integral dari $$$\frac{1}{u}$$$ adalah $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Ingat bahwa $$$u=\ln{\left(x \right)}$$$:

$$- \ln{\left(\left|{x - 1}\right| \right)} + \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = - \ln{\left(\left|{x - 1}\right| \right)} + \ln{\left(\left|{{\color{red}{\ln{\left(x \right)}}}}\right| \right)}$$

Oleh karena itu,

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

Tambahkan konstanta integrasi:

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

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