Integral dari $$$\frac{14}{\sqrt{3 - x}}$$$

Temukan $$$\int \frac{14}{\sqrt{3 - x}}\, dx$$$.

Terapkan aturan pengali konstanta $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ dengan $$$c=14$$$ dan $$$f{\left(x \right)} = \frac{1}{\sqrt{3 - x}}$$$:

$${\color{red}{\int{\frac{14}{\sqrt{3 - x}} d x}}} = {\color{red}{\left(14 \int{\frac{1}{\sqrt{3 - x}} d x}\right)}}$$

Misalkan $$$u=3 - x$$$.

Kemudian $$$du=\left(3 - x\right)^{\prime }dx = - dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dx = - du$$$.

Integralnya menjadi

$$14 {\color{red}{\int{\frac{1}{\sqrt{3 - x}} d x}}} = 14 {\color{red}{\int{\left(- \frac{1}{\sqrt{u}}\right)d u}}}$$

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

$$14 {\color{red}{\int{\left(- \frac{1}{\sqrt{u}}\right)d u}}} = 14 {\color{red}{\left(- \int{\frac{1}{\sqrt{u}} d u}\right)}}$$

Terapkan aturan pangkat $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ dengan $$$n=- \frac{1}{2}$$$:

$$- 14 {\color{red}{\int{\frac{1}{\sqrt{u}} d u}}}=- 14 {\color{red}{\int{u^{- \frac{1}{2}} d u}}}=- 14 {\color{red}{\frac{u^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}}}=- 14 {\color{red}{\left(2 u^{\frac{1}{2}}\right)}}=- 14 {\color{red}{\left(2 \sqrt{u}\right)}}$$

Ingat bahwa $$$u=3 - x$$$:

$$- 28 \sqrt{{\color{red}{u}}} = - 28 \sqrt{{\color{red}{\left(3 - x\right)}}}$$

Oleh karena itu,

$$\int{\frac{14}{\sqrt{3 - x}} d x} = - 28 \sqrt{3 - x}$$

Tambahkan konstanta integrasi:

$$\int{\frac{14}{\sqrt{3 - x}} d x} = - 28 \sqrt{3 - x}+C$$

$$$\int \frac{14}{\sqrt{3 - x}}\, dx = - 28 \sqrt{3 - x} + C$$$A