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

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

Integralkan suku demi suku:

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

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

$$- \int{\frac{1}{\sqrt{2 x - 1}} d x} + {\color{red}{\int{x^{2} d x}}}=- \int{\frac{1}{\sqrt{2 x - 1}} d x} + {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=- \int{\frac{1}{\sqrt{2 x - 1}} d x} + {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

Misalkan $$$u=2 x - 1$$$.

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

Integral tersebut dapat ditulis ulang sebagai

$$\frac{x^{3}}{3} - {\color{red}{\int{\frac{1}{\sqrt{2 x - 1}} d x}}} = \frac{x^{3}}{3} - {\color{red}{\int{\frac{1}{2 \sqrt{u}} d u}}}$$

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

$$\frac{x^{3}}{3} - {\color{red}{\int{\frac{1}{2 \sqrt{u}} d u}}} = \frac{x^{3}}{3} - {\color{red}{\left(\frac{\int{\frac{1}{\sqrt{u}} d u}}{2}\right)}}$$

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

$$\frac{x^{3}}{3} - \frac{{\color{red}{\int{\frac{1}{\sqrt{u}} d u}}}}{2}=\frac{x^{3}}{3} - \frac{{\color{red}{\int{u^{- \frac{1}{2}} d u}}}}{2}=\frac{x^{3}}{3} - \frac{{\color{red}{\frac{u^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}}}}{2}=\frac{x^{3}}{3} - \frac{{\color{red}{\left(2 u^{\frac{1}{2}}\right)}}}{2}=\frac{x^{3}}{3} - \frac{{\color{red}{\left(2 \sqrt{u}\right)}}}{2}$$

Ingat bahwa $$$u=2 x - 1$$$:

$$\frac{x^{3}}{3} - \sqrt{{\color{red}{u}}} = \frac{x^{3}}{3} - \sqrt{{\color{red}{\left(2 x - 1\right)}}}$$

Oleh karena itu,

$$\int{\left(x^{2} - \frac{1}{\sqrt{2 x - 1}}\right)d x} = \frac{x^{3}}{3} - \sqrt{2 x - 1}$$

Tambahkan konstanta integrasi:

$$\int{\left(x^{2} - \frac{1}{\sqrt{2 x - 1}}\right)d x} = \frac{x^{3}}{3} - \sqrt{2 x - 1}+C$$

$$$\int \left(x^{2} - \frac{1}{\sqrt{2 x - 1}}\right)\, dx = \left(\frac{x^{3}}{3} - \sqrt{2 x - 1}\right) + C$$$A