Integral dari $$$\frac{1}{\sqrt{x}}$$$

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

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

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

Oleh karena itu,

$$\int{\frac{1}{\sqrt{x}} d x} = 2 \sqrt{x}$$

Tambahkan konstanta integrasi:

$$\int{\frac{1}{\sqrt{x}} d x} = 2 \sqrt{x}+C$$

$$$\int \frac{1}{\sqrt{x}}\, dx = 2 \sqrt{x} + C$$$A