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

Temukan $$$\int \frac{1}{\sqrt{t}}\, dt$$$.

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

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

Oleh karena itu,

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

Tambahkan konstanta integrasi:

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

$$$\int \frac{1}{\sqrt{t}}\, dt = 2 \sqrt{t} + C$$$A