Integral dari $$$\frac{1}{f \left(2 a - x\right)}$$$ terhadap $$$x$$$

Temukan $$$\int \frac{1}{f \left(2 a - x\right)}\, dx$$$.

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

$${\color{red}{\int{\frac{1}{f \left(2 a - x\right)} d x}}} = {\color{red}{\frac{\int{\frac{1}{2 a - x} d x}}{f}}}$$

Misalkan $$$u=2 a - x$$$.

Kemudian $$$du=\left(2 a - x\right)^{\prime }dx = - dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dx = - du$$$.

Jadi,

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

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

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

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

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

Ingat bahwa $$$u=2 a - x$$$:

$$- \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{f} = - \frac{\ln{\left(\left|{{\color{red}{\left(2 a - x\right)}}}\right| \right)}}{f}$$

Oleh karena itu,

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

Tambahkan konstanta integrasi:

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

$$$\int \frac{1}{f \left(2 a - x\right)}\, dx = - \frac{\ln\left(\left|{2 a - x}\right|\right)}{f} + C$$$A