Integral dari $$$\frac{1}{- a + b}$$$ terhadap $$$a$$$

Temukan $$$\int \frac{1}{- a + b}\, da$$$.

Misalkan $$$u=- a + b$$$.

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

Dengan demikian,

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

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

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

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

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

Ingat bahwa $$$u=- a + b$$$:

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

Oleh karena itu,

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

Tambahkan konstanta integrasi:

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

$$$\int \frac{1}{- a + b}\, da = - \ln\left(\left|{a - b}\right|\right) + C$$$A