Integral dari $$$\frac{\ln\left(\sqrt{x}\right)}{x}$$$

Temukan $$$\int \frac{\ln\left(x\right)}{2 x}\, dx$$$.

Masukan ditulis ulang: $$$\int{\frac{\ln{\left(\sqrt{x} \right)}}{x} d x}=\int{\frac{\ln{\left(x \right)}}{2 x} d x}$$$.

Terapkan aturan pengali konstanta $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ dengan $$$c=\frac{1}{2}$$$ dan $$$f{\left(x \right)} = \frac{\ln{\left(x \right)}}{x}$$$:

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

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 $$$\frac{dx}{x} = du$$$.

Integral tersebut dapat ditulis ulang sebagai

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

Terapkan aturan pangkat $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ dengan $$$n=1$$$:

$$\frac{{\color{red}{\int{u d u}}}}{2}=\frac{{\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{2}=\frac{{\color{red}{\left(\frac{u^{2}}{2}\right)}}}{2}$$

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

$$\frac{{\color{red}{u}}^{2}}{4} = \frac{{\color{red}{\ln{\left(x \right)}}}^{2}}{4}$$

Oleh karena itu,

$$\int{\frac{\ln{\left(x \right)}}{2 x} d x} = \frac{\ln{\left(x \right)}^{2}}{4}$$

Tambahkan konstanta integrasi:

$$\int{\frac{\ln{\left(x \right)}}{2 x} d x} = \frac{\ln{\left(x \right)}^{2}}{4}+C$$

$$$\int \frac{\ln\left(x\right)}{2 x}\, dx = \frac{\ln^{2}\left(x\right)}{4} + C$$$A