Integral dari $$$- 12 \sqrt{x} - x^{2} + 1 - 2 e^{- 9 x}$$$

Temukan $$$\int \left(- 12 \sqrt{x} - x^{2} + 1 - 2 e^{- 9 x}\right)\, dx$$$.

Integralkan suku demi suku:

$${\color{red}{\int{\left(- 12 \sqrt{x} - x^{2} + 1 - 2 e^{- 9 x}\right)d x}}} = {\color{red}{\left(\int{1 d x} - \int{12 \sqrt{x} d x} - \int{x^{2} d x} - \int{2 e^{- 9 x} d x}\right)}}$$

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

$$- \int{12 \sqrt{x} d x} - \int{x^{2} d x} - \int{2 e^{- 9 x} d x} + {\color{red}{\int{1 d x}}} = - \int{12 \sqrt{x} d x} - \int{x^{2} d x} - \int{2 e^{- 9 x} d x} + {\color{red}{x}}$$

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

$$x - \int{12 \sqrt{x} d x} - \int{2 e^{- 9 x} d x} - {\color{red}{\int{x^{2} d x}}}=x - \int{12 \sqrt{x} d x} - \int{2 e^{- 9 x} d x} - {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=x - \int{12 \sqrt{x} d x} - \int{2 e^{- 9 x} d x} - {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ dengan $$$c=12$$$ dan $$$f{\left(x \right)} = \sqrt{x}$$$:

$$- \frac{x^{3}}{3} + x - \int{2 e^{- 9 x} d x} - {\color{red}{\int{12 \sqrt{x} d x}}} = - \frac{x^{3}}{3} + x - \int{2 e^{- 9 x} d x} - {\color{red}{\left(12 \int{\sqrt{x} d x}\right)}}$$

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

$$- \frac{x^{3}}{3} + x - \int{2 e^{- 9 x} d x} - 12 {\color{red}{\int{\sqrt{x} d x}}}=- \frac{x^{3}}{3} + x - \int{2 e^{- 9 x} d x} - 12 {\color{red}{\int{x^{\frac{1}{2}} d x}}}=- \frac{x^{3}}{3} + x - \int{2 e^{- 9 x} d x} - 12 {\color{red}{\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}=- \frac{x^{3}}{3} + x - \int{2 e^{- 9 x} d x} - 12 {\color{red}{\left(\frac{2 x^{\frac{3}{2}}}{3}\right)}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ dengan $$$c=2$$$ dan $$$f{\left(x \right)} = e^{- 9 x}$$$:

$$- 8 x^{\frac{3}{2}} - \frac{x^{3}}{3} + x - {\color{red}{\int{2 e^{- 9 x} d x}}} = - 8 x^{\frac{3}{2}} - \frac{x^{3}}{3} + x - {\color{red}{\left(2 \int{e^{- 9 x} d x}\right)}}$$

Misalkan $$$u=- 9 x$$$.

Kemudian $$$du=\left(- 9 x\right)^{\prime }dx = - 9 dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dx = - \frac{du}{9}$$$.

Integralnya menjadi

$$- 8 x^{\frac{3}{2}} - \frac{x^{3}}{3} + x - 2 {\color{red}{\int{e^{- 9 x} d x}}} = - 8 x^{\frac{3}{2}} - \frac{x^{3}}{3} + x - 2 {\color{red}{\int{\left(- \frac{e^{u}}{9}\right)d u}}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=- \frac{1}{9}$$$ dan $$$f{\left(u \right)} = e^{u}$$$:

$$- 8 x^{\frac{3}{2}} - \frac{x^{3}}{3} + x - 2 {\color{red}{\int{\left(- \frac{e^{u}}{9}\right)d u}}} = - 8 x^{\frac{3}{2}} - \frac{x^{3}}{3} + x - 2 {\color{red}{\left(- \frac{\int{e^{u} d u}}{9}\right)}}$$

Integral dari fungsi eksponensial adalah $$$\int{e^{u} d u} = e^{u}$$$:

$$- 8 x^{\frac{3}{2}} - \frac{x^{3}}{3} + x + \frac{2 {\color{red}{\int{e^{u} d u}}}}{9} = - 8 x^{\frac{3}{2}} - \frac{x^{3}}{3} + x + \frac{2 {\color{red}{e^{u}}}}{9}$$

Ingat bahwa $$$u=- 9 x$$$:

$$- 8 x^{\frac{3}{2}} - \frac{x^{3}}{3} + x + \frac{2 e^{{\color{red}{u}}}}{9} = - 8 x^{\frac{3}{2}} - \frac{x^{3}}{3} + x + \frac{2 e^{{\color{red}{\left(- 9 x\right)}}}}{9}$$

Oleh karena itu,

$$\int{\left(- 12 \sqrt{x} - x^{2} + 1 - 2 e^{- 9 x}\right)d x} = - 8 x^{\frac{3}{2}} - \frac{x^{3}}{3} + x + \frac{2 e^{- 9 x}}{9}$$

Tambahkan konstanta integrasi:

$$\int{\left(- 12 \sqrt{x} - x^{2} + 1 - 2 e^{- 9 x}\right)d x} = - 8 x^{\frac{3}{2}} - \frac{x^{3}}{3} + x + \frac{2 e^{- 9 x}}{9}+C$$

$$$\int \left(- 12 \sqrt{x} - x^{2} + 1 - 2 e^{- 9 x}\right)\, dx = \left(- 8 x^{\frac{3}{2}} - \frac{x^{3}}{3} + x + \frac{2 e^{- 9 x}}{9}\right) + C$$$A