Integral dari $$$- 2 x y + x$$$ terhadap $$$x$$$

Temukan $$$\int \left(- 2 x y + x\right)\, dx$$$.

Integralkan suku demi suku:

$${\color{red}{\int{\left(- 2 x y + x\right)d x}}} = {\color{red}{\left(\int{x d x} - \int{2 x y d x}\right)}}$$

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

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

Terapkan aturan pengali konstanta $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ dengan $$$c=2 y$$$ dan $$$f{\left(x \right)} = x$$$:

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

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

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

Oleh karena itu,

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

Sederhanakan:

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

Tambahkan konstanta integrasi:

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

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