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

Temukan $$$\int \frac{1}{a - b \sqrt{x}}\, dx$$$.

Misalkan $$$u=\sqrt{x}$$$.

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

Oleh karena itu,

$${\color{red}{\int{\frac{1}{a - b \sqrt{x}} d x}}} = {\color{red}{\int{\frac{2 u}{a - b u} d u}}}$$

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

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

Tulis ulang pembilang integran sebagai $$$ u =- \frac{1}{b}\left(- u b + a\right)+\frac{a}{b}$$$ dan pisahkan pecahannya:

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

Integralkan suku demi suku:

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

Terapkan aturan konstanta $$$\int c\, du = c u$$$ dengan $$$c=\frac{1}{b}$$$:

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

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

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

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

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

Oleh karena itu,

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

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

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

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

$$- \frac{2 a {\color{red}{\int{\frac{1}{v} d v}}}}{b^{2}} - \frac{2 u}{b} = - \frac{2 a {\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{b^{2}} - \frac{2 u}{b}$$

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

$$- \frac{2 a \ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{b^{2}} - \frac{2 u}{b} = - \frac{2 a \ln{\left(\left|{{\color{red}{\left(a - b u\right)}}}\right| \right)}}{b^{2}} - \frac{2 u}{b}$$

Ingat bahwa $$$u=\sqrt{x}$$$:

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

Oleh karena itu,

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

Sederhanakan:

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

Tambahkan konstanta integrasi:

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

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