Integral dari $$$x - \frac{7}{\sqrt{x}}$$$

Temukan $$$\int \left(x - \frac{7}{\sqrt{x}}\right)\, dx$$$.

Integralkan suku demi suku:

$${\color{red}{\int{\left(x - \frac{7}{\sqrt{x}}\right)d x}}} = {\color{red}{\left(- \int{\frac{7}{\sqrt{x}} d x} + \int{x d x}\right)}}$$

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

$$- \int{\frac{7}{\sqrt{x}} d x} + {\color{red}{\int{x d x}}}=- \int{\frac{7}{\sqrt{x}} d x} + {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=- \int{\frac{7}{\sqrt{x}} d x} + {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ dengan $$$c=7$$$ dan $$$f{\left(x \right)} = \frac{1}{\sqrt{x}}$$$:

$$\frac{x^{2}}{2} - {\color{red}{\int{\frac{7}{\sqrt{x}} d x}}} = \frac{x^{2}}{2} - {\color{red}{\left(7 \int{\frac{1}{\sqrt{x}} d x}\right)}}$$

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

$$\frac{x^{2}}{2} - 7 {\color{red}{\int{\frac{1}{\sqrt{x}} d x}}}=\frac{x^{2}}{2} - 7 {\color{red}{\int{x^{- \frac{1}{2}} d x}}}=\frac{x^{2}}{2} - 7 {\color{red}{\frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}}}=\frac{x^{2}}{2} - 7 {\color{red}{\left(2 x^{\frac{1}{2}}\right)}}=\frac{x^{2}}{2} - 7 {\color{red}{\left(2 \sqrt{x}\right)}}$$

Oleh karena itu,

$$\int{\left(x - \frac{7}{\sqrt{x}}\right)d x} = - 14 \sqrt{x} + \frac{x^{2}}{2}$$

Tambahkan konstanta integrasi:

$$\int{\left(x - \frac{7}{\sqrt{x}}\right)d x} = - 14 \sqrt{x} + \frac{x^{2}}{2}+C$$

$$$\int \left(x - \frac{7}{\sqrt{x}}\right)\, dx = \left(- 14 \sqrt{x} + \frac{x^{2}}{2}\right) + C$$$A