Integral dari $$$\frac{4 x}{\sqrt{x^{2} - 1}}$$$

Temukan $$$\int \frac{4 x}{\sqrt{x^{2} - 1}}\, dx$$$.

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

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

Jadi,

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

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=2$$$ dan $$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$:

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

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

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

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

$$4 \sqrt{{\color{red}{u}}} = 4 \sqrt{{\color{red}{\left(x^{2} - 1\right)}}}$$

Oleh karena itu,

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

Tambahkan konstanta integrasi:

$$\int{\frac{4 x}{\sqrt{x^{2} - 1}} d x} = 4 \sqrt{x^{2} - 1}+C$$

$$$\int \frac{4 x}{\sqrt{x^{2} - 1}}\, dx = 4 \sqrt{x^{2} - 1} + C$$$A