Integral dari $$$- 18 \sqrt{x} + x$$$

Temukan $$$\int \left(- 18 \sqrt{x} + x\right)\, dx$$$.

Integralkan suku demi suku:

$${\color{red}{\int{\left(- 18 \sqrt{x} + x\right)d x}}} = {\color{red}{\left(- \int{18 \sqrt{x} d x} + \int{x d x}\right)}}$$

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

$$- \int{18 \sqrt{x} d x} + {\color{red}{\int{x d x}}}=- \int{18 \sqrt{x} d x} + {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=- \int{18 \sqrt{x} d x} + {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ dengan $$$c=18$$$ dan $$$f{\left(x \right)} = \sqrt{x}$$$:

$$\frac{x^{2}}{2} - {\color{red}{\int{18 \sqrt{x} d x}}} = \frac{x^{2}}{2} - {\color{red}{\left(18 \int{\sqrt{x} d x}\right)}}$$

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

$$\frac{x^{2}}{2} - 18 {\color{red}{\int{\sqrt{x} d x}}}=\frac{x^{2}}{2} - 18 {\color{red}{\int{x^{\frac{1}{2}} d x}}}=\frac{x^{2}}{2} - 18 {\color{red}{\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}=\frac{x^{2}}{2} - 18 {\color{red}{\left(\frac{2 x^{\frac{3}{2}}}{3}\right)}}$$

Oleh karena itu,

$$\int{\left(- 18 \sqrt{x} + x\right)d x} = - 12 x^{\frac{3}{2}} + \frac{x^{2}}{2}$$

Tambahkan konstanta integrasi:

$$\int{\left(- 18 \sqrt{x} + x\right)d x} = - 12 x^{\frac{3}{2}} + \frac{x^{2}}{2}+C$$

$$$\int \left(- 18 \sqrt{x} + x\right)\, dx = \left(- 12 x^{\frac{3}{2}} + \frac{x^{2}}{2}\right) + C$$$A