Integral dari $$$- x e^{2} + 1$$$

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

Integralkan suku demi suku:

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

Terapkan aturan konstanta $$$\int c\, dx = c x$$$ dengan $$$c=1$$$:

$$- \int{x e^{2} d x} + {\color{red}{\int{1 d x}}} = - \int{x e^{2} d x} + {\color{red}{x}}$$

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

$$x - {\color{red}{\int{x e^{2} d x}}} = x - {\color{red}{e^{2} \int{x d x}}}$$

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

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

Oleh karena itu,

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

Sederhanakan:

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

Tambahkan konstanta integrasi:

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

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