Integral dari $$$\frac{1}{2 n - 1}$$$

Temukan $$$\int \frac{1}{2 n - 1}\, dn$$$.

Misalkan $$$u=2 n - 1$$$.

Kemudian $$$du=\left(2 n - 1\right)^{\prime }dn = 2 dn$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dn = \frac{du}{2}$$$.

Jadi,

$${\color{red}{\int{\frac{1}{2 n - 1} d n}}} = {\color{red}{\int{\frac{1}{2 u} d u}}}$$

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

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

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

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

Ingat bahwa $$$u=2 n - 1$$$:

$$\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} = \frac{\ln{\left(\left|{{\color{red}{\left(2 n - 1\right)}}}\right| \right)}}{2}$$

Oleh karena itu,

$$\int{\frac{1}{2 n - 1} d n} = \frac{\ln{\left(\left|{2 n - 1}\right| \right)}}{2}$$

Tambahkan konstanta integrasi:

$$\int{\frac{1}{2 n - 1} d n} = \frac{\ln{\left(\left|{2 n - 1}\right| \right)}}{2}+C$$

$$$\int \frac{1}{2 n - 1}\, dn = \frac{\ln\left(\left|{2 n - 1}\right|\right)}{2} + C$$$A