Integral dari $$$\frac{x}{\sqrt{2 x^{2} - 1}}$$$

Temukan $$$\int \frac{x}{\sqrt{2 x^{2} - 1}}\, dx$$$.

Misalkan $$$u=2 x^{2} - 1$$$.

Kemudian $$$du=\left(2 x^{2} - 1\right)^{\prime }dx = 4 x dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$x dx = \frac{du}{4}$$$.

Integralnya menjadi

$${\color{red}{\int{\frac{x}{\sqrt{2 x^{2} - 1}} d x}}} = {\color{red}{\int{\frac{1}{4 \sqrt{u}} d u}}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{4}$$$ dan $$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$:

$${\color{red}{\int{\frac{1}{4 \sqrt{u}} d u}}} = {\color{red}{\left(\frac{\int{\frac{1}{\sqrt{u}} d u}}{4}\right)}}$$

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

$$\frac{{\color{red}{\int{\frac{1}{\sqrt{u}} d u}}}}{4}=\frac{{\color{red}{\int{u^{- \frac{1}{2}} d u}}}}{4}=\frac{{\color{red}{\frac{u^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}}}}{4}=\frac{{\color{red}{\left(2 u^{\frac{1}{2}}\right)}}}{4}=\frac{{\color{red}{\left(2 \sqrt{u}\right)}}}{4}$$

Ingat bahwa $$$u=2 x^{2} - 1$$$:

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

Oleh karena itu,

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

Tambahkan konstanta integrasi:

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

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